Let the image of the point ((5)/(3),(5)/(3),(8)/(3)) in the plane x-2y+z-2=0 be P.

Question

Let the image of the point $(\frac{5}{3},\frac{5}{3},\frac{8}{3})$ 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

Image of point $(\frac{5}{3},\frac{5}{3},\frac{8}{3})$

$\begin{gathered} \frac{x-\frac{5}{3}}{1}=\frac{y-\frac{5}{3}}{-2}=\frac{z-\frac{8}{3}}{1}=\frac{-2(1	imes \frac{5}{3}+(-2)	imes \frac{8}{3}+1	imes \frac{8}{3}-2)}{{1}^{2}+{2}^{2}+{1}^{2}}=\frac{1}{3} \ 	herefore x=2,y=1,z=3 \ P{Q}^{2}={13}^{2}=(6-2{)}^{2}+(-2-1{)}^{2}+(\alpha -3{)}^{2} \ \Rightarrow (\alpha -3{)}^{2}=144\Rightarrow \alpha =15(\because \alpha >0) \end{gathered}$

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