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

The image of the point $\left(\frac{5}{3},\frac{5}{3},\frac{8}{3}\right)$ w.r.t plane will be,

$\frac{x-\frac{5}{3}}{1}=\frac{y-\frac{5}{3}}{-2}=\frac{2-\frac{8}{3}}{1}=\frac{-2\left(1\times \frac{5}{3}-2\times \frac{5}{3}+\frac{8}{3}-2\right)}{1^2+2^2+1^2}$

$\Rightarrow \frac{x-\frac{5}{3}}{1}=\frac{y-\frac{5}{3}}{-2}=\frac{2-\frac{8}{3}}{1}=\frac{-2(-1)}{6}$

$\Rightarrow P(x, y, z)=(2,1,3)$

But $PQ=13$

$4^2+3^2+(\alpha -3{)}^2=169$

$|\alpha -3|=12$

$\alpha =15$ Since $\left(\alpha >0\right)$

Hence this is the required answer.