The equation of the plane passing through the line of intersection of planes ${\pi }_1=2x+6y+4z-7=0$ , ${\pi }_2=x-y-2z-2=0$  and perpendicular to the plane $x+y+2z-5=0$ is

Equation of plane passes through the point of intersection of planes 
${\pi }_1=2x+6y+4z-7=0$   and ${\pi }_2=x-y-2z-2=0$  is
 $$\begin{gathered} (2x+6y+4z-7)+\lambda (x-y-2z-2)=0 \\ \Rightarrow (2+\lambda )x+(6-\lambda )y+(4-2\lambda )z-(7+2\lambda )=0\mathrm{............}\left(i\right) \end{gathered}$$
Plane (i) is perpendicular to plane
   $x+y+2z-5=0$
for some  $\lambda$, so
 $(2+\lambda )+(6-\lambda )+2(4-2\lambda )=0\Rightarrow 16=4\lambda \Rightarrow \lambda =4$
Therefore equation of required plane is
 $6x+2y-4z-15=0$