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

Given that $P_1:x-2y-2z+1=0$ $P_2:2x-3y-6z+1=0$ Equation of plane bisectors $\begin{array}{rl}& \left|\frac{x-2y-2z+1}{\sqrt{1+4+4}}\right|=\left|\frac{2x-3y-6z+1}{\sqrt{2^2+3^2+6^2}}\right|\\ & \frac{x-2y-2z+1}{3}=\pm \frac{2x-3y-6z+1}{7}\end{array}$ Since ${\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2=20>0$ $\therefore$ Negative sign will be taken for acute bisector. $\begin{array}{rl}& \Rightarrow 7x-14y-14z+7=-[6x-9y-18z+3]\\ & \Rightarrow 13x-23y-32z+10=0\\ & \left(-2,0,-\frac{1}{2}\right)\text{ satisfy it }\end{array}$