Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the required plane is $(x+2y+3z-4)+\lambda$ $$\begin{array}{rl}& (2x+y-z+5)=0\text{ i.e. }\\ & (1+2\lambda )x+(2+\lambda )y+(3-\lambda )z+(-4+5\lambda )=0\end{array}$$ Since (1) is perpendicular to the plane $5x+3y-6z+8=0$, we write $$\begin{array}{rl}& (1+2\lambda )(5)+(2+\lambda )(3)+(3-\lambda )(-6)=0\\ & \therefore 5+10\lambda +6+3\lambda -18+6\lambda =0\\ & 19\lambda =7\\ & \Rightarrow \lambda =\frac{7}{19}\end{array}$$ Substituting value of $\lambda$ in eq. (1), we get $$\begin{array}{rl}& \left(1+\frac{14}{19}\right)\mathrm{x}+\left(2+\frac{7}{19}\right)\mathrm{y}+\left(3-\frac{7}{19}\right)\mathrm{z}+\left(-4+\frac{35}{19}\right)=0\\ & \therefore 33\mathrm{x}+45\mathrm{y}+50\mathrm{z}-41=0\end{array}$$