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

3(1) + (-5) (3) + 2 (-$\alpha$ ) = 0

$3-15-2\alpha =0\Rightarrow \alpha +6=0\Rightarrow \alpha =-6$

Give the point (2, 1, -2) lie in the plane

$\therefore$  2 + 3 + 2$\alpha$ +$\beta$ =0

$5-12+\beta =0$

$\beta =7$

$\therefore \left(\alpha ,\beta \right)=\left(-6,7\right)$