Let the image of the point $\left(\frac{5}{3},\frac{5}{3}\text{,\hspace{0.17em}\hspace{0.17em}}\frac{8}{3}\right)$ in the plane $x-2y+z-2=0$  be P. If the distance of the point $Q\left(6,-2,\alpha \right),\alpha >0$  from P is 13 then  $\alpha =$

Use formula to find Image

$\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{l-z_1}{c}=\frac{-2\left(a{x}_1+b{y}_1+c{z}_1+d\right)}{a^2+b^2+c^2}$

And use distance formula

$\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2+{\left(z_1-z_2\right)}^2}$

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

$\text{ Image of point }\left(\frac{5}{3},\frac{5}{3},\frac{8}{3}\right)\text{is}\therefore \mathrm{x}=2,\mathrm{y}=1,\mathrm{z}=3$

$\begin{array}{r}{13}^2=(6-2{)}^2+(-2-1{)}^2+(\alpha -3{)}^2\\ \Rightarrow (\alpha -3{)}^2=144\Rightarrow \alpha =15(\because \alpha >0)\end{array}$