The equation of the plane through the intersection of the planes $3 x-y+2 z-4=0$ and $x+y+z-2=0$ and the point $(2,2,1)$ is

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 $(3 x-y+2 z-4)+\lambda(x+y+z-2)=0$...(i)
The plane passes through the point $(2,2,1)$. Therefore, this point will satisfy Eq. (i). $\begin{array}{cc} \therefore & (3 \times 2-2+2 \times 1-4)+\lambda(2+2+1-2)=0 \\ \Rightarrow & (6-4)+3 \lambda=0 \\ \Rightarrow & 2+3 \lambda=0 \\ \Rightarrow & \lambda=\frac{-2}{3} \end{array}$ On substituting this value of $\lambda$ in Eq. (i), we get the required plane as $\begin{array}{rr} (3 x-y+2 z-4)-\frac{2}{3}(x+y+z-2)=0 \\ \Rightarrow 9 x-3 y+6 z-12-2 x-2 y-2 z+4=0 \\ \Rightarrow 7 x-5 y+4 z-8=0 \end{array}$
This is the required equation of the plane.