Find the equation of the plane through the intersection of the planes

The equation of any plane through the intersection of the planes,

$3 x-y+2 z-4=0$ and $x+y+z-2=0$, is

$(3 x-y+2 z-4)+\alpha(x+y+z-2)=0$, where $\alpha \in \mathrm{R}$                 ...(1)

The plane passes through the point (2, 2, 1). Therefore, this point will satisfy equation (1).

$\therefore(3 \times 2-2+2 \times 1-4)+\alpha(2+2+1-2)=0$

$\Rightarrow 2+3 \alpha=0$

$\Rightarrow \alpha=-\frac{2}{3}$

Substituting $\alpha=-\frac{2}{3}$ in equation (1), we obtain

$(3 x-y+2 z-4)-\frac{2}{3}(x+y+z-2)=0$

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

$\Rightarrow(9 x-3 y+6 z-12)-2(x+y+z-2)=0$

$\Rightarrow 7 x-5 y+4 z-8=0$

This is the required equation of the plane.