The given planes are3x – y + 2z – 4 = 0 ...(1)and x + y + z – 2 = 0 ...(2)Any plane through the intersection of (1) and (2) is(3x – y + 2z – 4) + k (x + y + z – 2) = 0 ...(3)Since it passes through (2, 2, 1).∴ (6 – 2 + 2 – 4) + k (2 + 2 + 1 – 2) = 0$\Rightarrow 2+3k=0 \Rightarrow k=-\frac{2}{3}$Putting in (3), we get, $(3x - y + 2z - 4)$ $-\frac{2}{3}(x+y+z-2)=0$⇒ 9x – 3y + 6z – 12 – 2x – 2y – 2z + 4 = 0⇒ 7x – 5y + 4z – 8 = 0, which is the required equation.