Question

Let the line $L:\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $2x+y+3z=16$ 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 $PQR$, then ${\alpha }^2$ is equal to

Answer 1

Answer: 180

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 $QR=⟨2\lambda ,-\lambda ,\lambda +6⟩$
$DR$ of $L=⟨2,-1,1⟩$
$4\lambda +\lambda +\lambda +6=0$
$6\lambda +6=0\Rightarrow \lambda =-1$
$Q=(-1,0,2)$

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