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

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

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

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

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

$2\alpha -5=0\Rightarrow \alpha =\frac{5}{2}$

Also, the point $(3,-5,-2)$ on the line must satisfy the equation of the plane.

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

$3\alpha -13+\beta =0$

Substituting $\alpha =\frac{5}{2}$:

$3\left(\frac{5}{2}\right)-13+\beta =0$

$\frac{15}{2}-13+\beta =0$

$\beta =13-\frac{15}{2}=\frac{11}{2}$

Answer: $\frac{5}{2},\frac{11}{2}$