Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The reflection of plane ${\Pi }_1$ (normal $n_1=(2,3,4)$ ) in plane ${\Pi }_2$ (normal $n_2=(1,2,3)$ ) is $\left(n_2\cdot n_2\right){\Pi }_1-2\left(n_1\cdot n_2\right){\Pi }_2=0$. Here $n_2\cdot n_2=14$ and $n_1\cdot n_2=20$, so $14(2x+3y+4z+1)-40(x+2y+3z-2)=0$, i.e. $12x+38y+64z=94$, or $6x+19y+32z=47$.