$\text{ If the line }\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\text{ lie in the plane }x+3y-\alpha z+\beta =0\text{ then }\frac{1}{2}\left({\alpha }^2+{\beta }^2\right)=$

$\begin{array}{l}(2,1,-2)\text{ lies }x+3y-\alpha z+\beta =0\\ \Rightarrow 2\alpha +\beta =-5\\ \text{ Line lies in the plane }\\ \Rightarrow 3\left(1\right)+(-5)\left(3\right)+2(-\alpha )=0\Rightarrow 2\alpha =-12\\ \Rightarrow \alpha =-6,\beta =7\\ \frac{{\alpha }^2+{\beta }^2}{2}=\frac{36+49}{2}=42.5\end{array}$