Let Q be the mirror image of the point P(1,0,1) with respect to the plane S: x+y+z=5.

Question

Let $Q$ be the mirror image of the point $P(1,0,1)$ with respect to the plane $S:\ x+y+z=5$. If a line $L$ passing through $(1,-1,-1)$, parallel to the line $PQ$, meets the plane $S$ at $R$, then $QR^2$ is equal to:

Accepted Answer

Foot $M=(1+t,t,1+t)$ on $S$: $3t+2=5\Rightarrow t=1\Rightarrow M=(2,1,2)$.
$Q=2M-P=(3,2,3).$
$\vec{PQ}=(2,2,2)\propto(1,1,1)$. Line $L$: $(1+s,-1+s,-1+s)$; on $S$: $3s-1=5\Rightarrow s=2$.
$R=(3,1,1).$
$QR^2=(0)^2+(-1)^2+(-2)^2=5.$

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