Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Equation of plane passing through the intersection of given planes is $2x-y-4+\lambda (y+2z-4)=0$ Since, the plane passes through $(2,1,0)$ $\begin{array}{rl}& 2(2)-1-4+\lambda (1+2(0)-4)=0\\ & 4-1-4-3\lambda =0\\ & -1-3\lambda =0\\ & \lambda =\frac{-1}{3}\end{array}$ Substituting $\lambda =\frac{-1}{3}$ in equation (i), we get $\begin{array}{rl}& 2x-y-4-\frac{1}{3}(y+2z-4)=0\\ & 6x-3y-12-y-2z+4=0\\ & 6x-4y-2z-8=0\\ & 3x-2y-z=4\end{array}$