Let the image of the point (53,53,83) in the plane x-2y+z-2=0 be P.

Question

Let the image of the point $\left(\dfrac53,\dfrac53,\dfrac83\right)$ in the plane $x-2y+z-2=0$ be $P$. If the distance of the point $Q(6,-2,\alpha)$, $\alpha>0$, from $P$ is $13$, then $\alpha$ is equal to:

Answer: 15

Accepted Answer

$$\frac{x-\frac53}{1}=\frac{y-\frac53}{-2}=\frac{z-\frac83}{1}=\frac{-2\left(\frac53-\frac{10}{3}+\frac83-2\right)}{6}=\frac13.$$
$P=(2,1,3).$
$$13^2=(6-2)^2+(-2-1)^2+(\alpha-3)^2 \implies (\alpha-3)^2=144 \implies \alpha=15\ (\alpha>0).$$

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