Let a line pass through two distinct points P(-2,-1,3) and Q , and be parallel to the vector 3 i +2 j +2 k . If the distance of the point Q from the point R(1,3,3) is 5 , then the square of the area of P Q R is equal to

Correct answer: B. 136 Equation of line P Q is: (x+2)/(3)=(y+1)/(2)=(z-3)/(2)=r( say ) Let coordinate of Q=(3 r-2,2 r-1,2 r+3) P R=5 Then aligned & (3 r-2-1)^2+(2 r-1-3)^2+(2 r+3-3)^2=25 & r=0 or 2 & Coordinate of Q=(4,3,7) aligned aligned & square of area of P Q R=|(1)/(2)( P Q × P R )|^2 & =|(1)/(2)(6 i +4 j +4 k ) ×(3 i +4 j )|^2 & =|-8 i +6 j +6 k |^2=136 aligned

JEE Main 202522 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 22 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Three Dimensional Geometry

Let a line pass through two distinct points P(-2,-1,3) and Q , and...

Let a line pass through two distinct points

P(-2,-1,3)

and

Q

, and be parallel to the vector

3 \hat{i}+2 \hat{j}+2 \hat{k}

. If the distance of the point Q from the point

\mathrm{R}(1,3,3)

is 5 , then the square of the area of

\triangle P Q R

is equal to :

Options

A148
B136
C144
D140

Correct answer

B. 136

Step-by-step solution

Equation of line

P Q

is:

\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=r(\text{ say })

Let coordinate of

Q=(3 r-2,2 r-1,2 r+3) \because P R=5

Then

\begin{aligned} & (3 r-2-1)^2+(2 r-1-3)^2+(2 r+3-3)^2=25 \\ & \therefore r=0 \text{\text{ or } } 2 \\ & \therefore \quad \text{ Coordinate of } Q=(4,3,7) \end{aligned}

\begin{aligned} \therefore & \text{ square of area of } \triangle P Q R=\left|\frac{1}{2}(\overrightarrow{P Q} \times \overrightarrow{P R})\right|^2 \\ & =\left|\frac{1}{2}(6 \hat{i}+4 \hat{j}+4 \hat{k}) \times(3 \hat{i}+4 \hat{j})\right|^2 \\ & =|-8 \hat{i}+6 \hat{j}+6 \hat{k}|^2=136\end{aligned}

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