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

$\because$ The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $$x+3y-\alpha z+\beta =0$$ $\therefore \mathrm{Pt}(2,1,-2)$ lies on the plane i.e. $2+3+$ $2\alpha +\beta =0$ or $2\alpha +\beta +5=0$ $\dots .(\mathrm{i})$ Also normal to plane will be perpendicular to line, $\therefore 3\times 1-5\times 3+2\times (-\alpha )=0\Rightarrow \alpha =-6$ From equation (i) then, $\beta =7$ $\therefore (\alpha ,\beta )=(-6,7)$