Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The given line is $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ and the plane is $x+3y-\alpha z+\beta =0$.

Since the line lies in the plane, the direction vector of the line is perpendicular to the normal vector of the plane. Therefore, their dot product is zero.

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

$3-15-2\alpha =0$

$-12-2\alpha =0\Rightarrow \alpha =-6$

Also, any point on the line must satisfy the equation of the plane. A point on the line is $(2,1,-2)$. Substituting this point into the plane equation, we get:

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

$5+2\alpha +\beta =0$

Substituting $\alpha =-6$:

$5+2(-6)+\beta =0$

$5-12+\beta =0\Rightarrow \beta =7$

Thus, $(\alpha ,\beta )=(-6,7)$.

Answer: $(-6,7)$