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 :

• A

  148
• Correct answer
  B

  136
• C

  144
• D

  140

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 $\begin{aligned} & (3 r-2-1)^2+(2 r-1-3)^2+(2 r+3-3)^2=25 \\ & \therefore r=0 \text { or } 2 \\ & \therefore \quad \text { Coordinate of } Q=(4,3,7) \end{aligned}$
$\begin{aligned} \therefore & \text { square of area of } \triangle P Q R=\left|\frac{1}{2}(\overrightarrow{P Q} \times \overrightarrow{P R})\right|^2 \\ & =\left|\frac{1}{2}(6 \hat{i}+4 \hat{j}+4 \hat{k}) \times(3 \hat{i}+4 \hat{j})\right|^2 \\ & =|-8 \hat{i}+6 \hat{j}+6 \hat{k}|^2=136\end{aligned}$

Hence, the correct answer is option: 'B'.