Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$  lie in the plane $x+3y-\alpha z+\beta =0$  then $(\alpha ,\beta )=$  
 

Since the line is perpendicular to the plane
a1 a2+ b1 b2+ c1 c2= 0
3(1)+(-5)(3)+2(- $\alpha$) = 0
3-15-2 $\alpha$= 0
-12 = 2$\alpha$ 
$\alpha$= -6
(2,1,-2) lies on x+3y-$\alpha$ z+$\beta$ = 0
2+3(1)-(-6) )(-2)+$\beta$  = 0
2+3-12+ $\beta$ = 0
5-12+ $\beta$ = 0
-7+  $\beta$= 0
$\beta$= 7
$\left(\alpha ,\beta \right)$= (-6,7)