Question

Suppose the line $\frac{x-2}{\alpha }=\frac{y-2}{-5}=\frac{z+2}{2}$ lies on the plane $x+3y-2z+\beta =0$. Then $(\alpha +\beta )$ is equal to

Answer 1

Line $\frac{x-2}{\alpha }=\frac{y-2}{-5}=\frac{z+2}{2}$ lies in $x+3y-2z+\beta =0$.
Direction ratios of normal vector of plane $=(1,3,-2)$
Angle between line and plane $=0$
Direction ratios of line $=(\alpha ,-5,2)$
Dot product of above direction ratios $=0$
$\left(1\right)\cdot \left(\alpha \right)+(-5)\cdot \left(3\right)+\left(2\right)\cdot (-2)=0$
$\Rightarrow \alpha =19$
Point $(2,2,-2)$ lies on plane
$\left(1\right)\left(2\right)+3\left(2\right)-2(-2)+\beta =0$
$\beta =-12$
$\alpha +\beta =19-12=7$