Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Equation of plane through the intersection of given planes is $\begin{array}{rl}& (x+y+z-1)+\lambda (2x+3y-z+4)=0\dots \text{ (i) }\\ & \Rightarrow (1+2\lambda )x+(1+3\lambda )y+(1-\lambda )z+4\lambda -1=0\end{array}$ Since the plane is parallel to $\mathrm{Y}$-axis. $\begin{array}{rl}\therefore & 1+3\lambda =0\\ & \Rightarrow \lambda =\frac{-1}{3}\end{array}$ Substituting $\lambda =\frac{-1}{3}$ in (i), we get $\begin{array}{rl}& (x+y+z-1)-\frac{1}{3}(2x+3y-z+4)=0\\ & \Rightarrow x+4z-7=0\end{array}$ Point $(3,2,1)$ satisfies this equation.