Let the line L: x-12=y+1-1=z-31 intersect the plane 2x+y+3z=16 at the point P.

Question

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

Answer: 180

Accepted Answer

General point on $L$: $(2\lambda+1,\ -\lambda-1,\ \lambda+3)$. On the plane:
$$2(2\lambda+1)+(-\lambda-1)+3(\lambda+3)=16 \implies 6\lambda+10=16 \implies \lambda=1.$$
$P=(3,-2,4).$
Foot of perpendicular $Q$ from $R$: $\vec{QR}=(2\lambda,-\lambda,\lambda+6)$, perpendicular to $(2,-1,1)$:
$$4\lambda+\lambda+\lambda+6=0 \implies 6\lambda+6=0 \implies \lambda=-1.$$
$Q=(-1,0,2).$
$$\vec{QR}=(2,-1,-5),\qquad \vec{QP}=(4,-2,2).$$
$$\vec{QR}\times\vec{QP}=(-12,-24,0).$$
$$\alpha=\frac12\sqrt{144+576} \implies \alpha^2=\frac{720}{4}=180.$$

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