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 $R(1,3,3)$ is $5$, then the square of the area of $\triangle PQR$ is equal to:
Step-by-Step Solution
Key Concept: Write $Q=P+r(3,2,2)$ (since $PQ$ is parallel to $(3,2,2)$), apply $|QR|=5$ to find $r$, then compute the square of the area using $\left|\tfrac{1}{2}\overrightarrow{PQ}\times\overrightarrow{PR}\right|^2$.
$Q=(3r-2,2r-1,2r+3)$. $|QR|^2=(3r-3)^2+(2r-4)^2+(2r)^2=25$.
$9r^2-18r+9+4r^2-16r+16+4r^2=25 \Rightarrow 17r^2-34r=0 \Rightarrow r=0$ or $r=2$.
Take $r=2$: $Q=(4,3,7)$.
$\overrightarrow{PQ}=(6,4,4)$, $\overrightarrow{PR}=(3,4,0)$.
$\overrightarrow{PQ}\times\overrightarrow{PR}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\6&4&4\\3&4&0\end{vmatrix}=-16\hat{i}+12\hat{j}+12\hat{k}$.
$\text{Area}^2=\left(\tfrac{1}{2}\right)^2|{-16\hat{i}+12\hat{j}+12\hat{k}}|^2=\tfrac{1}{4}(256+144+144)=\tfrac{544}{4}=136$.
Correct Answer: 2