3D Geometry Class 12 JEE Main Questions: Weightage, PYQs, and Formulas

Weightage of 3D Geometry in JEE Main 2026

Based on the 2026 exam analysis, 3D Geometry contributes 6.11% weightage in the JEE Main maths paper. In practical terms, 2-3 questions appear in most of the shifts of the exam and around 29 overall in both sessions, worth roughly 8 to 12 marks out of the 100 marks allotted to Maths.

Topic-Wise Weightage and Number of Questions

The table below shows the topic-wise distribution of questions from the Three-Dimensional Geometry chapter in the January and April sessions of the JEE Main 2026 exam.

Also Read: JEE Mains 2027: Maths Set of 5 Sample Paper with Solution

Last 10 Years' Weightage of Three-Dimensional Geometry in JEE Main

The 3D geometry chapter question has appeared constantly over the past 10 years and has remained an important topic of JEE Main, with questions frequently appearing from Equations of a line, shortest distance between two lines, image of a point, and perpendicular distance.

Important Formulas of Three-Dimensional Geometry for JEE Main

1. Direction Cosines (1, m, n) of a line: $1^2+m^2+n^2=1$

2. Direction Ratios (a, b, c) to Direction Cosines: $1=\mathrm{a} / \sqrt{ }\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right), \mathrm{m}=\mathrm{b} / \sqrt{ }\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right), \mathrm{n}=$ $\mathrm{c} / \sqrt{ }\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)$

3. Equation of a line through a point $\left(x_1, y_1, z_1\right)$ with direction ratios $(a, b, c):\left(x-x_1\right) / a=$ $\left(\mathrm{y}-\mathrm{y}_1\right) / \mathrm{b}=\left(\mathrm{z}-\mathrm{z}_1\right) / \mathrm{c}$

4. Angle between two lines with direction ratios $\left(\mathbf{a}_1, \mathbf{b}_1, \mathbf{c}_1\right)$ and $\left(\mathbf{a}_2, \mathbf{b}_2, \mathbf{c}_2\right): \cos \theta=\mid \mathbf{a}_1 \mathbf{a}_2+$ $\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2 / /\left[\sqrt{ }\left(\mathrm{a}_1{ }^2+\mathrm{b}_1{ }^2+\mathrm{c}_1{ }^2\right) \cdot \sqrt{ }\left(\mathrm{a}_2{ }^2+\mathrm{b}_2{ }^2+\mathrm{c}_2{ }^2\right)\right]$

5. Shortest distance between two skew lines (using position vectors and direction vectors): $d=\left|\left(b_1 \times b_2\right) \cdot\left(a_2-a_1\right)\right| /\left|b_1 \times b_2\right|$

6. Equation of a plane (normal form): $\mathrm{ax}+\mathrm{by}+\mathrm{cz}=\mathrm{d}$, where (a, b, c) are direction ratios of the normal

7. Distance of point $\left(\mathbf{x}_{\mathbf{1}}, \mathbf{y}_{\mathbf{1}}, \mathbf{z}_{\mathbf{1}}\right)$ from plane $\mathbf{a x}+\mathbf{b y}+\mathbf{c z}+\mathbf{d}=\mathbf{0}: \mathbf{D}=\left|\mathrm{ax}_1+\mathrm{by}_1+\mathrm{cz}_1+\mathrm{d}\right| /$ $\sqrt{ }\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)$

8. Angle between two planes with normals $\left(\mathbf{a}_{\mathbf{1}}, \mathbf{b}_{\mathbf{1}}, \mathbf{c}_{\mathbf{1}}\right)$ and $\left(\mathbf{a}_{\mathbf{2}}, \mathbf{b}_{\mathbf{2}}, \mathbf{c}_{\mathbf{2}}\right): \cos \theta=\mid \mathrm{a}_{\mathbf{1}} \mathrm{a}_{\mathbf{2}}+\mathrm{b}_{\mathbf{1}} \mathrm{b}_2+$ $\mathrm{c}_1 \mathrm{c}_2 \mid /\left[\sqrt{ }\left(\mathrm{a}_1{ }^2+\mathrm{b}_1{ }^2+\mathrm{c}_1{ }^2\right) \cdot \sqrt{ }\left(\mathrm{a}_2{ }^2+\mathrm{b}_2{ }^2+\mathrm{c}_2{ }^2\right)\right]$

9. Condition for coplanarity of two lines through points A and B with direction vectors $\mathrm{b}_1$ and $\mathrm{b}_2:(\mathrm{AB}) \cdot\left(\mathrm{b}_1 \times \mathrm{b}_2\right)=0$

3D Geometry Practice Questions for JEE Main

Question 1:Let $\mathrm{L}_1: \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}-1}{-1}=\frac{\mathrm{z}+1}{0}$ and $\mathrm{L}_2: \frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}}{0}=\frac{\mathrm{z}+4}{\alpha}, \alpha \in \mathrm{R}$, be two lines, which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $\mathrm{A}(1,1,-1)$ on $\mathrm{L}_2$, then the value of $26 \alpha(\mathrm{~PB})^2$ is_____________.

(1) 216

(2) -

(3) -

(4) -

Solution: Option (1)

We are given two lines: $

L_1: \frac{x - 1}{3} = \frac{y - 1}{-1} = \frac{z + 1}{0}, \quad

L_2: \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{\alpha}

$

Let the parameter for

$ L_1 $ be $ \lambda $.

Then a point on $ L_1 $ is given by: $

x = 3\lambda + 1, \quad y = -\lambda + 1, \quad z = -1

$ Let the parameter for $ L_2 $ be $ \mu $.

Then a point on $ L_2 $ is: $

x = 2\mu + 2, \quad y = 0, \quad z = \alpha \mu - 4

$

Equating the coordinates (since lines intersect), we get: $

3\lambda + 1 = 2\mu + 2 \quad \text{(1)}

$

$

-\lambda + 1 = 0 \quad \text{(2)}

$

$

-1 = \alpha \mu - 4 \quad \text{(3)}

$

From equation (2): $

\lambda = 1

$ Substitute into equation (1): $

3(1) + 1 = 2\mu + 2 \Rightarrow 4 = 2\mu + 2 \Rightarrow \mu = 1

$ Substitute into equation (3): $

-1 = \alpha(1) - 4 \Rightarrow \alpha = 3

$

So, the point of intersection $ B $ is: $

x = 2(1) + 2 = 4, \quad y = 0, \quad z = 3(1) - 4 = -1

\Rightarrow B = (4, 0, -1)

$

Now, find the foot of the perpendicular from point $ A(1,1,-1) $ to line $ L_2 $.

Let a general point on $ L_2 $ be given by a parameter $ \delta $: $

P = (2\delta + 2, 0, 3\delta - 4)

$

Then, vector $ \vec{AP} $ is: $

\vec{AP} = \langle 2\delta + 2 - 1, \ 0 - 1, \ 3\delta - 4 + 1 \rangle

= \langle 2\delta + 1, -1, 3\delta - 3 \rangle

$

The direction vector of $ L_2 $ is $ \langle 2, 0, 3 \rangle $. Since $ \vec{AP} $ is perpendicular to the line, their dot product is zero: $

(2\delta + 1)(2) + (-1)(0) + (3\delta - 3)(3) = 0

\Rightarrow 4\delta + 2 + 9\delta - 9 = 0

\Rightarrow 13\delta - 7 = 0 \Rightarrow \delta = \frac{7}{13}

$ Substitute into $ P $: $

x = 2\cdot \frac{7}{13} + 2 = \frac{14}{13} + \frac{26}{13} = \frac{40}{13}

$

$y = 0$

$

z = 3\cdot \frac{7}{13} - 4 = \frac{21}{13} - \frac{52}{13} = \frac{-31}{13}

$ So, the foot of the perpendicular is: $

P = \left( \frac{40}{13}, 0, \frac{-31}{13} \right)

$

Now compute $ PB^2 $, where $ B = (4, 0, -1) $: $

PB^2 = \left(4 - \frac{40}{13} \right)^2 + (0 - 0)^2 + \left(-1 - \left(\frac{-31}{13}\right)\right)^2

= \left( \frac{12}{13} \right)^2 + \left( \frac{18}{13} \right)^2

$ $

PB^2 = \frac{144}{169} + \frac{324}{169} = \frac{468}{169}

$ Given $ \alpha = 3 $, we calculate: $

26 \cdot \alpha \cdot PB^2 = 26 \cdot 3 \cdot \frac{468}{169}

= \frac{36504}{169} = 216

$

Hence, the answer is 216.

Question 2: Let $P$ be the foot of the perpendicular from the point $\mathrm{Q}(10,-3,-1)$ on the line $\frac{\mathrm{x}-3}{7}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}+1}{-2}$. Then the area of the right angled triangle $P Q R$, where $R$ is the point $(3,-2,1)$, is

(1) $9 \sqrt{15}$

(2) $\sqrt{30}$

(3) $8\sqrt{15}$

(4) $3\sqrt{30}$

Solution: Option (4)

Given: Point $ Q(10, -3, -1) $ and line

$

\frac{x - 3}{7} = \frac{y - 2}{-1} = \frac{z + 1}{-2}.

$ Let this common ratio be $ \lambda $. Then the coordinates of a general point $ P $ on the line are:

$

P = (3 + 7\lambda,\ 2 - \lambda,\ -1 - 2\lambda).

$ Now, vector $ \overrightarrow{PQ} = \vec{Q} - \vec{P} $:

$

\overrightarrow{PQ} = \left(10 - (3 + 7\lambda),\ -3 - (2 - \lambda),\ -1 - (-1 - 2\lambda)\right)

= (7 - 7\lambda,\ -5 + \lambda,\ 2\lambda).

$ Direction vector of the line is $ \vec{d} = \langle 7, -1, -2 \rangle $. Since $ \overrightarrow{PQ} \perp \vec{d} $, their dot product is zero:

$

(7 - 7\lambda)(7) + (-5 + \lambda)(-1) + (2\lambda)(-2) = 0

$

$

49 - 49\lambda + 5 - \lambda - 4\lambda = 0

$

$

(49 + 5) - (49\lambda + \lambda + 4\lambda) = 0

\Rightarrow 54 - 54\lambda = 0

\Rightarrow \lambda = 1

$ Substitute $ \lambda = 1 $ into point $ P $:

$

P = (3 + 7, 2 - 1, -1 - 2) = (10, 1, -3)

$ So, coordinates of $ P = (10, 1, -3) $, $ Q = (10, -3, -1) $, $ R = (3, -2, 1) $.

Now find vectors:

$

\overrightarrow{PQ} = Q - P = (0, -4, 2), \quad \overrightarrow{PR} = R - P = (-7, -3, 4)

$

Now use the cross product to find the area:

$

\vec{A} = \overrightarrow{PQ} \times \overrightarrow{PR}

=

\begin{vmatrix}

\hat{i} & \hat{j} & \hat{k} \\

0 & -4 & 2 \\

-7 & -3 & 4

\end{vmatrix}

= \hat{i}( -4 \cdot 4 - 2 \cdot (-3)) - \hat{j}(0 \cdot 4 - 2 \cdot (-7)) + \hat{k}(0 \cdot (-3) - (-4) \cdot (-7))

$

$

= \hat{i}(-16 + 6) - \hat{j}(0 + 14) + \hat{k}(0 - 28)

= \langle -10, -14, -28 \rangle

$ Magnitude of cross product:

$

|\vec{A}| = \sqrt{(-10)^2 + (-14)^2 + (-28)^2} = \sqrt{100 + 196 + 784} = \sqrt{1080}

$ Area of triangle:

$

\text{Area} = \frac{1}{2} \cdot |\vec{A}| = \frac{1}{2} \sqrt{1080} = \frac{\sqrt{1080}}{2}

$ Simplify:

$

\sqrt{1080} = \sqrt{36 \cdot 30} = 6\sqrt{30}

\Rightarrow \text{Area} = \frac{6\sqrt{30}}{2} = 3\sqrt{30} \text{ sq. units}

$

Hence, the answer is option (4).

Question 3:Let the circle C touch the line $\mathrm{x}-\mathrm{y}+1=0$, have the centre on the positive $x$-axis, and cut off a chord of length $\frac{4}{\sqrt{13}}$ along the line $-3x+2y=1$. Let H be the hyperbola $\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1$, whose one of the foci is the centre of C and the length of the transverse axis is the diameter of C . Then $2 \alpha^2+3 \beta^2$ is equal to_______________.

(1) 19

(2) -

(3) -

(4) -

Solution: Option (1)

Let the center of the circle $C$ be at $(a,0)$, since it lies on the positive $x$-axis. Let the radius be $r$. The circle touches the line $x - y + 1 = 0$. The distance from the centre to this line must be equal to the radius:

$

\frac{|a - 0 + 1|}{\sqrt{1^2 + (-1)^2}} = r \Rightarrow \frac{|a + 1|}{\sqrt{2}} = r \Rightarrow r = \frac{a + 1}{\sqrt{2}} \quad \text{(1)}

$ Now, the circle cuts a chord of length $\frac{4}{\sqrt{13}}$ along the line $-3x + 2y = 1$.

Let the perpendicular distance from the centre $(a, 0)$ to this line be $d$. Use the perpendicular distance formula:

$

d = \frac{|-3a + 2 \cdot 0 - 1|}{\sqrt{(-3)^2 + 2^2}} = \frac{| -3a - 1 |}{\sqrt{13}} = \frac{3a + 1}{\sqrt{13}} \quad \text{(2)}

$ Let $2l = \frac{4}{\sqrt{13}}$ be the length of the chord, then $l = \frac{2}{\sqrt{13}}$. The perpendicular from the centre to the chord bisects it. Using the right triangle formed by the radius, the half-chord, and the perpendicular, apply the Pythagorean theorem:

$

r^2 = d^2 + l^2 = \left( \frac{3a + 1}{\sqrt{13}} \right)^2 + \left( \frac{2}{\sqrt{13}} \right)^2

\Rightarrow r^2 = \frac{(3a + 1)^2 + 4}{13} \quad \text{(3)}

$ Now substitute $r = \frac{a + 1}{\sqrt{2}}$ from equation (1):

$

\left( \frac{a + 1}{\sqrt{2}} \right)^2 = \frac{(3a + 1)^2 + 4}{13}

\Rightarrow \frac{(a + 1)^2}{2} = \frac{(3a + 1)^2 + 4}{13}

$ Multiply both sides by 26:

$

13(a + 1)^2 = 2[(3a + 1)^2 + 4]

\Rightarrow 13(a^2 + 2a + 1) = 2(9a^2 + 6a + 1 + 4)

\Rightarrow 13a^2 + 26a + 13 = 18a^2 + 12a + 10

\Rightarrow -5a^2 + 14a + 3 = 0

$ Solve the quadratic:

$

5a^2 - 14a - 3 = 0

\Rightarrow a = \frac{14 \pm \sqrt{196 + 60}}{10} = \frac{14 \pm \sqrt{256}}{10} = \frac{14 \pm 16}{10}

\Rightarrow a = 3 \text{ or } a = -\frac{1}{5}

$ Since $a$ is on the positive $x$-axis, $a = 3$. Now, the radius is:

$

r = \frac{a + 1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}

\Rightarrow \text{Diameter} = 2r = 4\sqrt{2}

$

The centre of the hyperbola is at the origin, and one focus is at $(3, 0)$, so $c = 3$ (distance from centre to focus).

Also, the transverse axis has length $2\alpha = 4\sqrt{2}$, so $\alpha = 2\sqrt{2}$:

$

\Rightarrow \alpha^2 = (2\sqrt{2})^2 = 8

$ In a hyperbola, $c^2 = \alpha^2 + \beta^2$:

$

9 = 8 + \beta^2 \Rightarrow \beta^2 = 1

$ Finally,

$

2\alpha^2 + 3\beta^2 = 2 \cdot 8 + 3 \cdot 1 = 16 + 3 = 19

$

Hence, the answer is 19.

Question 4: Let P be the image of the point $\mathrm{Q}(7,-2,5)$ in the line $L: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $R(5, p, q)$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is_____________.

(1) 957

(2) -

(3) -

(4) -

Solution: Option (1)

We are given point $ Q = (7, -2, 5) $, and line $ L: \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4} $. Let point $ R = (5, p, q) $ lie on the line. We are to find the square of the area of triangle $ \triangle PQR $, where $ P $ is the image of $ Q $ in line $ L $. Let a general point on line $ L $ be:

$

R(\lambda) = (2\lambda + 1, 3\lambda - 1, 4\lambda)

$ Since $ R = (5, p, q) $, equating:

$

2\lambda + 1 = 5 \Rightarrow \lambda = 2

\Rightarrow R = (5, 5, 8)

$ Let the foot of perpendicular from $ Q $ to line $ L $ be $ T = (2\lambda + 1, 3\lambda - 1, 4\lambda) $ Now, vector $ \vec{QT} = (2\lambda - 6)\hat{i} + (3\lambda + 1)\hat{j} + (4\lambda - 5)\hat{k} $ Direction vector of line $ L $ is $ \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} $ Using perpendicularity condition:

$

\vec{QT} \cdot \vec{b} = 0

$

$

(2\lambda - 6)(2) + (3\lambda + 1)(3) + (4\lambda - 5)(4) = 0

$

$

4\lambda - 12 + 9\lambda + 3 + 16\lambda - 20 = 0

\Rightarrow 29\lambda - 29 = 0 \Rightarrow \lambda = 1

$ Substitute $ \lambda = 1 $ to get:

$

T = (3, 2, 4)

$ Now, $ P $ is the reflection of $ Q $ in point $ T $:

$

P = 2T - Q = (6 - 7, 4 - (-2), 8 - 5) = (-1, 6, 3)

$ Now, find lengths $ QT $ and $ RT $: $

|QT| = \sqrt{(3 - 7)^2 + (2 + 2)^2 + (4 - 5)^2} = \sqrt{16 + 16 + 1} = \sqrt{33}

$ $

|RT| = \sqrt{(5 - 3)^2 + (5 - 2)^2 + (8 - 4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}

$ Since $ QT \perp RT $, area of triangle is:

ar($\triangle QRT$)$ = \frac{1}{2} \cdot |QT| \cdot |RT| = \frac{1}{2} \cdot \sqrt{33} \cdot \sqrt{29}

$ Since P is the image of Q, then the areas above and below RT will be equal. And the $ar(\triangle PQR)=2ar(\triangle QTR)$ $ar(\triangle PQR)=2 \times \frac{1}{2} \cdot \sqrt{33} \cdot \sqrt{29}$ $ar(\triangle PQR)= \sqrt{33} \cdot \sqrt{29}=\sqrt957$ $

\Rightarrow \text{Area}^2 = \left(\frac{1}{2} \cdot \sqrt{957}\right)^2 = 957

$

Hence, the answer is 957.

Question 5: If the image of the point $(4,4,3)$ in the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to:

(1) 9

(2) 12

(3) 8

(4) 7

Solution: Option (1)

Given the point $P = (4,4,3)$ and the line

$\frac{x-1}{2} = \frac{y-2}{1} = \frac{z-1}{3}$

let the image of $P$ in the line be $R(\alpha, \beta, \gamma)$.

The direction vector of the line is

$

\vec{d} = 2\hat{i} + \hat{j} + 3\hat{k}.

$ Let the foot of the perpendicular from $P$ on the line be

$

Q = (1 + 2\lambda, 2 + \lambda, 1 + 3\lambda).

$ Since $\overrightarrow{P Q}$ is perpendicular to $\vec{d}$, we have:

$

\overrightarrow{P Q} \cdot \vec{d} = 0,

$

where

$

\overrightarrow{P Q} = (1 + 2\lambda - 4,\, 2 + \lambda - 4,\, 1 + 3\lambda - 3) = (2\lambda - 3,\, \lambda - 2,\, 3\lambda - 2).

$

Calculate the dot product:

$

2(2\lambda - 3) + 1(\lambda - 2) + 3(3\lambda - 2) = 0,

$

which simplifies to:

$

4\lambda - 6 + \lambda - 2 + 9\lambda - 6 = 0,

$

$

14 \lambda - 14 = 0 \implies \lambda = 1.

$ Substitute $\lambda = 1$ back into $Q$:

$

Q = (1 + 2 \times 1,\, 2 + 1,\, 1 + 3 \times 1) = (3,3,4).

$

Since $R$ is the image of $P$ in the line with foot $Q$, $Q$ is the midpoint of segment $P R$, so:

$

\frac{\alpha + 4}{2} = 3, \quad \frac{\beta + 4}{2} = 3, \quad \frac{\gamma + 3}{2} = 4.

$ Solving for $\alpha, \beta, \gamma$:

$

\alpha + 4 = 6 \implies \alpha = 2,

$

$

\beta + 4 = 6 \implies \beta = 2,

$

$

\gamma + 3 = 8 \implies \gamma = 5.

$ Therefore,

$

\alpha + \beta + \gamma = 2 + 2 + 5 = 9.

$

Hence, the correct answer is option (1).

Question 6: If $\alpha=1 + \sum_{r=1}^6 {(-3)^{r-1}}.{^{12} \mathrm{C}_{2 \mathrm{r}-1}}$, then the distance of the point $(12, \sqrt{3})$ from the line $\alpha x-\sqrt{3} y+1=0$ is_____________.

(1) 5

(2) -

(3) -

(4) -

Solution: Option (1)

We are given:

$\alpha = 1 + \sum_{r=1}^6 (-3)^{r-1} \cdot {^{12}C_{2r - 1}}$ Let us simplify the sum:

This is a sum over all odd binomial coefficients of $^{12}C_k$ multiplied by powers of $-3$ starting from 0. Let us define the sum:

$\sum_{r=1}^6 (-3)^{r-1} \cdot {^{12}C_{2r - 1}} = S$ Let’s compute $S = \sum_{r=1}^6 (-3)^{r-1} \cdot {^{12}C_{2r - 1}}$ Compute each term: $r = 1$: $(-3)^0 \cdot {^{12}C_1} = 1 \cdot 12 = 12$

$r = 2$: $(-3)^1 \cdot {^{12}C_3} = (-3) \cdot 220 = -660$

$r = 3$: $(-3)^2 \cdot {^{12}C_5} = 9 \cdot 792 = 7128$

$r = 4$: $(-3)^3 \cdot {^{12}C_7} = (-27) \cdot 792 = -21384$

$r = 5$: $(-3)^4 \cdot {^{12}C_9} = 81 \cdot 220 = 17820$

$r = 6$: $(-3)^5 \cdot {^{12}C_{11}} = (-243) \cdot 12 = -2916$ Now sum all:

$S = 12 - 660 + 7128 - 21384 + 17820 - 2916 = 0$ So $\alpha = 1 + S = 1 + 0 = 1$ Now find the distance of point $(12, \sqrt{3})$ from line: $\alpha x - \sqrt{3} y + 1 = 0 \Rightarrow x - \sqrt{3} y + 1 = 0$ Distance from point $(x_0, y_0)$ to line $ax + by + c = 0$ is:

$d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$ Here, $a = 1$, $b = -\sqrt{3}$, $c = 1$, $(x_0, y_0) = (12, \sqrt{3})$ $d = \frac{|1 \cdot 12 + (-\sqrt{3}) \cdot \sqrt{3} + 1|}{\sqrt{1^2 + (\sqrt{3})^2}} = \frac{|12 - 3 + 1|}{\sqrt{1 + 3}} = \frac{10}{2} = 5$

Hence, the answer is 5.

Question 7: Let $P$ be the foot of the perpendicular from the point $(1,2,2)$ on the line $L: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$. Let the line $\overrightarrow{\mathrm{r}}=(-\hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}), \lambda \in \mathbf{R}$, intersect the line L at Q . Then $2(\mathrm{PQ})^2$ is equal to:

(1) 27

(2) 25

(3) 29

(4) 19

Solution: Option (1)

We are given:

- A point $ A = (1, 2, 2) $

- A line $ L: \frac{x - 1}{1} = \frac{y + 1}{-1} = \frac{z - 2}{2} = \mu $

- A second line: $ \vec{r} = (-\hat{i} + \hat{j} - 2\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) $ Find the foot of perpendicular $ P $ from point $ A $ to line $ L $ Let a general point on $ L $ be:

$

P(\mu + 1, -\mu - 1, 2\mu + 2)

$ The direction vector of line $ L $ is $ \vec{d} = \langle 1, -1, 2 \rangle $ Vector $ \vec{AP} = \langle \mu + 1 - 1, -\mu - 1 - 2, 2\mu + 2 - 2 \rangle = \langle \mu, -\mu - 3, 2\mu \rangle $ Since $ \vec{AP} $ is perpendicular to direction vector $ \vec{d} $, their dot product must be zero: $

\vec{AP} \cdot \vec{d} = \mu(1) + (-\mu - 3)(-1) + 2\mu(2) = \mu + \mu + 3 + 4\mu = 6\mu + 3 = 0

\Rightarrow \mu = -\frac{1}{2}

$ Now substitute $ \mu = -\frac{1}{2} $ into the coordinates of $ P $:

$

P = \left( \frac{-1}{2} + 1, -\left( -\frac{1}{2} \right) - 1, 2 \cdot \left( -\frac{1}{2} \right) + 2 \right)

= \left( \frac{1}{2}, -\frac{1}{2}, 1 \right)

$ Find point $ Q $, where the second line intersects line $ L $ General point on second line:

$

Q = (-1 + \lambda, 1 - \lambda, -2 + \lambda)

$ Let this point also lie on line $ L $, whose general point is:

$

(x, y, z) = (\mu + 1, -\mu - 1, 2\mu + 2)

$ Equating:

$

\mu + 1 = -1 + \lambda \Rightarrow \mu = \lambda - 2 \tag{1}

$

$

-\mu - 1 = 1 - \lambda \Rightarrow \mu = \lambda - 2 \tag{2}

$

$

2\mu + 2 = -2 + \lambda \Rightarrow 2(\lambda - 2) + 2 = -2 + \lambda

$

$

2\lambda - 4 + 2 = -2 + \lambda \Rightarrow 2\lambda - 2 = -2 + \lambda

\Rightarrow \lambda = 0

\Rightarrow \mu = -2

$ Substitute back:

$

Q = (-1 + 0, 1 - 0, -2 + 0) = (-1, 1, -2)

$ Compute the distance $ PQ $ $

\vec{PQ} = \langle -1 - \tfrac{1}{2}, 1 - (-\tfrac{1}{2}), -2 - 1 \rangle

= \left\langle -\tfrac{3}{2}, \tfrac{3}{2}, -3 \right\rangle

$ $

PQ^2 = \left( \tfrac{3}{2} \right)^2 + \left( \tfrac{3}{2} \right)^2 + 9 = \tfrac{9}{4} + \tfrac{9}{4} + 9 = \tfrac{18}{4} + 9 = \tfrac{54}{4}

$ $

2(PQ)^2 = 2 \cdot \frac{54}{4} = 27

$

Hence, the correct answer is option (1).

Question 8: Consider the lines $\mathrm{x}(3 \lambda+1)+\mathrm{y}(7 \lambda+2)=17 \lambda+5$, $\lambda$ being a parameter, all passing through a point P . One of these lines (say L) is farthest from the origin. If the distance of L from the point $(3,6)$ is $d$, then the value of $d^2$ is

(1) 20

(2) 30

(3) 10

(4) 15

Solution: Option (1)

$x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$

$\Rightarrow 3 x \lambda+x+7 y \lambda+2 y=17 \lambda+5$

$\Rightarrow \lambda(3 x+7 y-17)+(x+2 y-5)=0$ For the equation to hold for all $\lambda$, the coefficients of $\lambda$ and the constant term must be zero

$3 x+7 y-17=0$

$x+2 y-5=0$ $-3 x-6 y+15=0$

$y-2=0$, so $y=2$

Substitute $y=2$

$x+2 y-5=0\Rightarrow x+4-5=0$, so $x=1$ The point $P$ is $(1,2)$. The line farthest from the origin is perpendicular to the line connecting the origin and $P$.

The slope of the line connecting the origin and P is $\frac{2-0}{1-0}=2$.

The slope of the perpendicular line L is $-\frac{1}{2}$.

The equation of line $L$ passing through $(1,2)$ with slope $-\frac{1}{2}$ is:

$y-2=-\frac{1}{2}(x-1)$

$2y-4=-x+1$

$x+2 y-5=0$ The distance from $(3,6)$ to $x+2 y-5=0$

$d=\frac{|1(3)+2(6)-5|}{\sqrt{1^2+2^2}}$

$d=\frac{|3+12-5|}{\sqrt{5}}$

$d=\frac{10}{\sqrt{5}}$

$

\begin{aligned}

d^2 & =\left(\frac{10}{\sqrt{5}}\right)^2 \\

d^2 & =\frac{100}{5} \\

d^2 & =20

\end{aligned}

$

Hence, the answer is option (1).

Question 9: The shortest distance between the curves $\mathrm{y}^2=8 \mathrm{x}$ and $x^2+y^2+12 y+35=0$ is :

(1) $2 \sqrt{3}-1$

(2) $\sqrt{2}$

(3) $3 \sqrt{2}-1$

(4) $2 \sqrt{2}-1$

Solution: Option (4)

Equation of normal: $y=m x-2 a m-a m^3 \quad(a=2)$ $y=m x-4 m-2 m^3$ Centre of circle: $C(0,-6)$, radius $=1$ $\begin{aligned}

& -6=-4 m-2 m^3 \\

& \Rightarrow m=1 \\

& P\left(a m^2,-2 a m\right) \\

& =P(2,-4)

\end{aligned}$ Shortest distance: $CP - r$ $\begin{aligned}

& =\sqrt{4+4}-1 \\

& =2 \sqrt{2}-1 \text{ units}

\end{aligned}$

Hence, the correct answer is option (4).

Question 10: The distance of the point $(7,10,11)$ from the line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ along the line $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$ is

(1) 18

(2) 14

(3) 12

(4) 16

Solution: Option (2)

Equation of line passing through $P(7,10,11)$ along the line $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$ is $\frac{x-7}{2}=\frac{y-10}{3}=\frac{z-11}{6}=\lambda$ Let the point on the line be $Q(2 \lambda+7,3 \lambda+10,6 \lambda+11)$ $Q$ lies on line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ $\begin{aligned}

& 3 \lambda+10=4 \Rightarrow \lambda=-2 \\

& \therefore \quad Q(3,4,-1) \\

& P Q=\sqrt{16+36+144}=14 \text{ units}

\end{aligned}$

Hence, the correct answer is option (2).

Question 11: If the image of the point $\mathrm{P}(1,0,3)$ in the line joining the points $\mathrm{A}(4,7,1)$ and $\mathrm{B}(3,5,3)$ is $Q(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to

(1) $\frac{47}{3}$

(2) $\frac{46}{3}$

(3) 18

(4) 13

Solution: Option (2)

Given: $P(1, 0, 3), \quad A(4, 7, 1), \quad B(3, 5, 3)$ Direction ratios of line $AB = B - A = (-1, -2, 2)$ Parametric form of line $AB:

\quad x = \lambda + 3, \quad y = 2\lambda + 5, \quad z = -2\lambda + 3$ Let foot of perpendicular from P to line AB be $R(\lambda + 3, 2\lambda + 5, -2\lambda + 3)$ Vector $\overrightarrow{PR} = \langle \lambda + 2, 2\lambda + 5, -2\lambda \rangle$ Direction vector of line $AB = \langle 1, 2, -2 \rangle$ Dot product must be zero: $\overrightarrow{PR} \cdot \vec{d}_{AB} = 0$ $\Rightarrow (\lambda + 2)(1) + (2\lambda + 5)(2) + (-2\lambda)(-2) = 0$ $\Rightarrow \lambda + 2 + 4\lambda + 10 + 4\lambda = 0$ $\Rightarrow 9\lambda + 12 = 0 \Rightarrow \lambda = -\frac{4}{3}$ Coordinates of foot $R:

\quad x = \frac{5}{3}, \quad y = \frac{7}{3}, \quad z = \frac{17}{3}$ Image of P about point R is: $Q = 2R - P = \left( \frac{10}{3} - 1, \frac{14}{3} - 0, \frac{34}{3} - 3 \right) = \left( \frac{7}{3}, \frac{14}{3}, \frac{25}{3} \right)$ $\Rightarrow \alpha + \beta + \gamma = \frac{7 + 14 + 25}{3} = \frac{46}{3}$

Hence, the correct answer is option (2).

Question 12: Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

(1) 42

(2) 46

(3) 48

(4) 40

Solution: Option (2)

$\mathrm{A}(3, \alpha, 3)$ and $B(-3,-7, \beta)$ Given lines:

$

L_1: \frac{x-3}{3} = \frac{y-\alpha}{-1} = \frac{z-3}{1}

$

$

L_2: \frac{x+3}{-3} = \frac{y+7}{2} = \frac{z-\beta}{4}

$ Direction vectors:

$

\vec{d_1} = \langle 3, -1, 1 \rangle, \quad \vec{d_2} = \langle -3, 2, 4 \rangle

$ Points on each line:

$

\vec{P_1} = (3, \alpha, 3), \quad \vec{P_2} = (-3, -7, \beta)$ Shortest distance between skew lines: $

D = \frac{|(\vec{P_2} - \vec{P_1}) \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}

$ Calculate $\vec{P_2} - \vec{P_1}$:

$

\vec{P_2} - \vec{P_1} = (-3 - 3, -7 - \alpha, \beta - 3) = (-6, -7 - \alpha, \beta - 3)

$ Calculate $\vec{d_1} \times \vec{d_2}$:

$

\vec{d_1} \times \vec{d_2} =

\begin{vmatrix}

\mathbf{i} & \mathbf{j} & \mathbf{k} \\

3 & -1 & 1 \\

-3 & 2 & 4 \\

\end{vmatrix}

= \mathbf{i}((-1)(4) - 1 \cdot 2) - \mathbf{j}(3 \cdot 4 - 1 \cdot (-3)) + \mathbf{k}(3 \cdot 2 - (-1)(-3))

$ $

= \mathbf{i}(-4 - 2) - \mathbf{j}(12 + 3) + \mathbf{k}(6 - 3)

= \mathbf{i}(-6) - \mathbf{j}(15) + \mathbf{k}(3)

= \langle -6, -15, 3 \rangle

$ Magnitude of $\vec{d_1} \times \vec{d_2}$:

$

|\vec{d_1} \times \vec{d_2}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3 \sqrt{30}

$ Calculate the numerator $|(\vec{P_2} - \vec{P_1}) \cdot (\vec{d_1} \times \vec{d_2})|:$

$

= |(-6)(-6) + (-7 - \alpha)(-15) + (\beta - 3)(3)|

= |36 + 15(7 + \alpha) + 3(\beta - 3)|

$ $

= |36 + 105 + 15 \alpha + 3 \beta - 9|

= |132 + 15 \alpha + 3 \beta|

$ Given shortest distance $D = 3 \sqrt{30}$, so:

$

D = \frac{|132 + 15 \alpha + 3 \beta|}{3 \sqrt{30}} = 3 \sqrt{30}

$ Multiply both sides by $3 \sqrt{30}$:

$

|132 + 15 \alpha + 3 \beta| = 9 \times 30 = 270

$ Divide both sides by 3:

$

|44 + 5 \alpha + \beta| = 90

$ So,

$

44 + 5 \alpha + \beta = \pm 90

$ Solve for $5\alpha + \beta$: 1. If

$

44 + 5 \alpha + \beta = 90 \implies 5 \alpha + \beta = 46

$ 2. If

$

44 + 5 \alpha + \beta = -90 \implies 5 \alpha + \beta = -134

$ Positive value of $5 \alpha + \beta$ is $46$.

Hence, the correct answer is option (2).

Question 13: The line $L_{1}$ is parallel to the vector $\vec{a}=-3 \hat{i}+2 \hat{j}+4 \hat{k}$ and passes through the point $(7,6,2)$ and the line $L_{2}$ is parallel to the vector $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ and passes through the point $(5,3,4)$. The shortest distance between the lines $L_{1}$ and $L_{2}$ is :

(1) $\frac{23}{\sqrt{38}}$

(2) $\frac{21}{\sqrt{57}}$

(3) $\frac{23}{\sqrt{57}}$

(4) $\frac{21}{\sqrt{38}}$

Solution: Option (1)

Given: Line $L_1$ passes through the point $B(7,6,2)$ and is parallel to the vector $\vec{a} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ Line $L_2$ passes through the point $A(5,3,4)$ and is parallel to the vector $\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}$ To find: The shortest distance between lines $L_1$ and $L_2$. Use the formula, $

\text{Distance} = \frac{|(\vec{b} \times \vec{a}) \cdot \vec{AB}|}{|\vec{b} \times \vec{a}|}

$ Find $\vec{AB}$ $

\vec{AB} = \vec{B}-\vec{A} = (7 - 5)\hat{i} + (6 - 3)\hat{j} + (2 - 4)\hat{k} = 2\hat{i} + 3\hat{j} - 2\hat{k}

$ Calculate $\vec{b} \times \vec{a}$ $

\vec{b} \times \vec{a} =

\begin{vmatrix}

\hat{i} & \hat{j} & \hat{k} \\

2 & 1 & 3 \\

-3 & 2 & 4

\end{vmatrix}

= \hat{i}(1 \cdot 4 - 3 \cdot 2) - \hat{j}(2 \cdot 4 - 3 \cdot (-3)) + \hat{k}(2 \cdot 2 - 1 \cdot (-3))

$ $

= \hat{i}(4 - 6) - \hat{j}(8 + 9) + \hat{k}(4 + 3) = -2\hat{i} -17\hat{j} + 7\hat{k}

$ Similarly, find the dot product $(\vec{b} \times \vec{a}) \cdot \vec{AB}$ $

(-2\hat{i} -17\hat{j} + 7\hat{k}) \cdot (2\hat{i} + 3\hat{j} - 2\hat{k}) = (-2)(2) + (-17)(3) + (7)(-2)

$ $

= -4 - 51 - 14 = -69

$ Now find, $|\vec{b} \times \vec{a}|$ $

|\vec{b} \times \vec{a}| = \sqrt{(-2)^2 + (-17)^2 + 7^2} = \sqrt{4 + 289 + 49} = \sqrt{342}

$ Calculate the shortest distance $

\text{Distance} = \frac{|-69|}{\sqrt{342}} = \frac{69}{\sqrt{342}}

$ On simplifying, $

\text{Distance} = \frac{23}{\sqrt{38}}\text{ units}

$

Hence, the correct answer is option (1).

Question 14: Let the line L pass through $(1,1,1)$ and intersect the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}$ Then, which of the following points lies on the line L?

(1) $(4,22,7)$

(2) $(5,4,3)$

(3) $(10,-29,-50)$

(4) $(7,15,13)$

Solution: Option (4)

Dr's of $\mathrm{AC} \Rightarrow 2 \lambda, 3 \lambda-2,4 \lambda$

Dr's of BC $\Rightarrow \mu+2,2 \mu+3, \mu-1$

$\Rightarrow \frac{\mu+2}{2 \lambda}=\frac{2 \mu+3}{3 \lambda-2}=\frac{\mu-1}{4 \lambda}$

$\Rightarrow 2(\mu+2)=\mu-1 \Rightarrow \mu=-5$

$\Rightarrow$ Dr's of $\mathrm{BC} \Rightarrow 3,7,6$

$\Rightarrow$ equation of $L \Rightarrow \frac{x-1}{3}=\frac{y-1}{7}=\frac{z-1}{6}$

$(7,15,13)$ satisfies.

Hence, the correct answer is option (4).

Question 15: If the equation of the line passing through the point $\left(0,-\frac{1}{2}, 0\right)$ and perpendicular to the lines $\overrightarrow{\mathrm{r}}=\lambda(\hat{\mathrm{i}}+a \hat{\mathrm{j}}+b \hat{\mathrm{k}})$ and $\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}-6 \hat{\mathrm{k}})+\mu(-b \hat{\mathrm{i}}+a \hat{\mathrm{j}}+5 \hat{\mathrm{k}})$ is $\frac{x-1}{-2}=\frac{y+4}{d}=\frac{z-c}{-4}$, then $a+b+c+d$ is equal to :

(1) 10

(2) 14

(3) 13

(4) 12

Solution: Option (2)

Direction ratios of the given line are $-2, d$, and -4 $\Rightarrow-2+a d-4 b=0$ and $2 b+a d-20=0$

Subtracting equations (i) and (ii) $\begin{aligned}

& 6 b-18=0 \\

& b=3

\end{aligned}$ also, line passes through the point $\left(0, \frac{-1}{2}, 0\right)$ $\begin{aligned}

& \Rightarrow \frac{0-1}{-2}=\frac{-\frac{1}{2}+4}{d}=\frac{0-c}{-4} \\

& \Rightarrow d=7, c=2

\end{aligned}$ From equation (i) $\begin{aligned}

& -2+7 a-12=0 \\

& \Rightarrow a=2 \\

& a+b+c+d=14

\end{aligned}$

Hence, the correct answer is option (2).