A plane which bisects the angle between the two given planes$2 x-y+2z-4=0$ and $x+2y+2z-2=0,$ passes through the point

 

Solution:

 Given planes are $2x-y+2z-4=0$ and 

$x+2y+2z-2=0$

Equation of angle bisectors are

$\begin{array}{r}\frac{2x-y+2z-4}{\sqrt{4+1+4}}=\pm \left(\frac{x+2y+2z-2}{\sqrt{1+4+4}}\right)\\ \Rightarrow \frac{2x-y+2z-4}{3}=\pm \left(\frac{x+2y+2z-2}{3}\right)\end{array}$

$\therefore 2x-y-2z-4=x+2y+2z-2\Rightarrow x-3y-2=0$

or $2x-y+2z-4=-(x+2y+2z-2)$

$\Rightarrow 3x+y+4z-6=0$

Since, Point $(2,-4,1)$ satisfies equation (ii).

So, required point is $(2,-4,1)$.