Question

Let the line $L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $2 x+y+3 z=16$ at the point $P$. Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$. If $\alpha$ is the area of triangle $P Q R$, then $\alpha^2$ is equal to

Answer 1

Any point on $L ((2 \lambda+1),(-\lambda-1),(\lambda+3))$
$ 2(2 \lambda+1)+(-\lambda-1)+3(\lambda+3)=16$
$6 \lambda+10=16 \Rightarrow \lambda=1 $
$ \therefore P =(3,-2,4)$
$DR$ of $QR =\langle 2 \lambda,-\lambda, \lambda+6\rangle$
$DR$ of $L =\langle 2,-1,1\rangle$
$4 \lambda+\lambda+\lambda+6=0 $
$ 6 \lambda+6=0 \Rightarrow \lambda=-1 $
$Q=(-1,0,2)$

$\overrightarrow{ QR } \times \overrightarrow{ QP }=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & -1 & -5 \\ 4 & -2 & 2\end{vmatrix}=-12 \hat{ i }-24 \hat{ j }$
$\alpha=\frac{1}{2} \times \sqrt{144+576} \Rightarrow \alpha^2=\frac{720}{4}=180$