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

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

The direction ratios of the line are $(3,-5,2)$ and the normal to the plane has direction ratios $(1,3,\alpha )$.

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

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

$3-15+2\alpha =0$

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

Also, the point $(2,1,-2)$ on the line must satisfy the equation of the plane.

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

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

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

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

Answer: $(6,7)$