Mathematics - Vector Question with Solution | TestHub

Equation of line $PQ$ is: $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=r(\text{ say })$ Let coordinate of $Q=(3r-2,2r-1,2r+3)$ $\because PR=5$ Then $\begin{array}{rl}& (3r-2-1{)}^2+(2r-1-3{)}^2+(2r+3-3{)}^2=25\\ & \therefore r=0\text{ or }2\\ & \therefore \text{ Coordinate of }Q=(4,3,7)\end{array}$
$\begin{array}{rl}\therefore & \text{ square of area of }△PQR={\left|\frac{1}{2}(\overrightarrow{PQ}\times \overrightarrow{PR})\right|}^2\\ & ={\left|\frac{1}{2}(6\hat{i}+4\hat{j}+4\hat{k})\times (3\hat{i}+4\hat{j})\right|}^2\\ & =|-8\hat{i}+6\hat{j}+6\hat{k}{|}^2=136\end{array}$