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

The given planes are $3x - y + 2z - 4 = 0$ ..(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 $(2,2,1)$
$\therefore$ $(6 - 2 + 2 - 4) + k(2 + 2 + 1 - 2) = 0$

$\Rightarrow$ $2 + 3k = 0 \Rightarrow k = - \cfrac{2}{3}$

Putting in (3), $(3x - y + 2z - 4) - \cfrac{2}{3}(x + y + z - 2) = 0$

$\Rightarrow$ $9x - 3y + 6z - 12 - 2x - 2y - 2z + 4 = 0$

$\Rightarrow$ $7x - 5y + 4z - 8 = 0$ ,which is the required equation.