Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Equation of plane passing through the line of intersection of given planes is $(x+2y-3z+2)+\lambda (6x+y+z+1)=0$ $(1+6\lambda )x+(2+\lambda )y+(-3+\lambda )z+(2+\lambda )=0$ This is parallel to the line $\frac{x-1}{1}=\frac{y+2}{1}=\frac{z-7}{-1}$ $\therefore (1+6\lambda )(1)+(2+\lambda )(1)+(-3+\lambda )(-1)=0\Rightarrow \lambda =-1$ Hence required equation of plane is $-5x+y-4z+1=0\Rightarrow 5x-y+4z=1$