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

line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta =0$ $\therefore (3)(1)+(-5)(3)+2(-\alpha )=0\Rightarrow \alpha =-6$ Thus equation of plane is $x+3y+6z+\beta =0$ and point $(2,1,-2)$ lies in it. $$\begin{array}{rl}& \therefore 2+3(1)+6(-2)+\beta =0\Rightarrow \beta =7\\ & \therefore \alpha \beta =(-6)(7)=-42\end{array}$$