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

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) 195
3) 224
4) 167

Correct answer: (1)

Solution:

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
$

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 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) 7
3) 23
4) 11

Correct answer: (1)

Solution:

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 center 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 center $(a, 0)$ 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 center 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 center of the hyperbola is at the origin, and one focus is at $(3, 0)$, so $c = 3$ (distance from center 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 3: 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) 876
3) 245
4) 765

Correct answer: (1)

Solution:

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 4: 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) 2
3) 9
4) 10

Correct answer: (1)

Solution:

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 5: 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

Correct answer: (1)

Solution:

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 6: 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

Correct answer: (2)

Solution:

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).

Question 7: 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

Correct answer: (2)

Solution:

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 8: 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) $(7,15,13)$

Correct answer: (4)

Solution:

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 9: 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

Correct answer: (1)

Solution:

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 10: 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) $\frac{21}{\sqrt{38}}$

Correct answer: (1)

Solution:

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 11: 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) $2 \sqrt{2}-1$

Correct answer: (4)

Solution:

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 12: 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

Correct answer: (2)

Solution:

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 13: 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

Correct answer: (2)

Solution:

$\mathrm{A}(3, \alpha, 3)$ 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 14: 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

Correct answer: (1)

Solution:

$x(3 \lambda+1)+y(7 \lambda+2)=17 \lambda+5$
$\Rightarrow 3 x \lambda+x+7 y \lambda+2 y=17 \lambda+5$
$\Rightarrow \lambda(3 x+7 y-17)+(x+2 y-5)=0$

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 15: 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) $3\sqrt{30}$

Correct answer: (4)

Solution:

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 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 the option (4).