Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Equation of the plane passing through the line of intersection of the given planes, is $(xz+2y+3z-4)+\lambda (4x+3y+2z-1)=0$
Since the plane (1) passes through origin, We get $-4-\lambda =0\Rightarrow \lambda =-4$ Substituting value of $\lambda$ in equation (1), we get $$\begin{array}{rl}& -15x-10y-5z=0\Rightarrow 3x+2y+z=0\\ & \therefore \text{ d.r.s. are }(3,2,1)\end{array}$$