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

Required equation of plane is  ${\pi }_1+\lambda {\pi }_2=0$

$\begin{array}{l}(2x+6y+4z-7)+\lambda (x-y-2z-2)=0\\ \Rightarrow (2+\lambda )x+(6-\lambda )y+(4-2\lambda )z-7-2\lambda \mathrm{.............}\left(1\right)\\ \left(1\right){\perp }^{r}x+y+2z-5=0\\ \therefore (2+\lambda )+(6-\lambda )+2(4-2\lambda )=0\\ \Rightarrow \lambda =4\\ \lambda =4in\left(1\right)\Rightarrow 6x+2y-4z-15=0\end{array}$