Let the line $\frac{\mathrm{x}-2}{3}=\frac{\mathrm{y}-1}{-5}=\frac{\mathrm{z}+2}{2}$ lie in the plane $\mathrm{x}+3\mathrm{y}-\mathrm{\alpha z}+\mathrm{\beta }=0\text{. Then }{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=\mathrm{\_\_\_\_\_\_}$

The d.r.’s of the line are (3, –5, 2).

$\Rightarrow$The d.r,’s of the normal to the plane are $(1,3,-\mathrm{\alpha }).$

Since line lies on the plane, the line is perpendicular
to the normal of the plane.

$$\begin{gathered} \Rightarrow 3\left(1\right)+(-5)3+2(-\alpha )=0 \\ \Rightarrow 3-15-2\alpha =0\Rightarrow \alpha =-6 \\ (2,1,-2) lies in x+3y-\alpha z+\beta =0 \\ \Rightarrow 2+3+6(-2)+\beta =0 \\ \Rightarrow \beta =7 \\ {\alpha }^2+{\beta }^2=36+49=85 \end{gathered}$$