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, $(\alpha , \beta )$

equals

(-6, 7)

Given line passes through (2, 1, -2) and is parallel to $\vec{b} = 3\hat{i} - 5\hat{j} + 2\hat{k}.$ A vector normal to the given plane is $\vec{n} = \hat{i} + 3\hat{j} - \alpha \hat{k}.$

If the line lies in the plane, then (2, 1, -2) lies on the plane and $\vec{b}$ is normal to $\vec{n}$.

$∴ 2 + 3 + 2\alpha + \beta = 0 $ and $ 3-15 - 2\alpha = 0 $

$⇒ 5 + 2 \alpha + \beta = 0 $ and $ \alpha = -6 ⇒ \alpha = -6 $ and $\beta = 7 $