Determine if two given planes are parallel and find points on their intersection

Let $P = (2,3,1), \ A = (1,2,3), \ B = (3,2,1)$ and let $M = \{ P + sA + tB \}$ . Let $M'$ be the plane determined by the Cartesian equation $x - 2y + z = 0$ .

Determine if the two planes are parallel.
If we define $M''$ to be the plane with Cartesian equation $x+2y+z = 0$ find two points on $M' \cap M''$ .

We know for any point $Q \notin M$ , there is a unique plane parallel to $M$ containing $Q$ . We pick a point $Q$ on $M'$ that is not on $M$ and show that the unique plane parallel to $M$ containing $A$ is, in fact, $M'$ .
The point $Q = (0,0,0)$ is on $M'$ since

$(0) - 2(0) + 0 = 0.$

It is not on $M$ since

$\begin{align*} 2+s+3t &= 0 \\ 3+2s + 2t &= 0 \\ 1+3s + t &= 0 \end{align*}$

has no solution $s,t$ . So, the unique plane parallel to $M$ containing $(0,0,0)$ is

$\{ (0,0,0) + sA + tB \} = \{ (s+3t, 2s+2t, 3s+t) \}.$

Then, we obtain the Cartesian equation of this plane. We have

$\begin{align*} s + 3t &=x \\ 2s+2t &= y \\ 3s + t &= z. \end{align*}$

This gives us $s = -3t + x$ . Then, $y = -6t + 2x + 2t$ which implies

$t = \frac{2x-y}{4}, \qquad \implies \qquad s = -3 \left( \frac{2x-y}{4} \right) + x.$

Hence,

$\begin{align*} z &= -9 \left( \frac{2x-y}{4} \right) + 3x + \left( \frac{2x-y}{4} \right) \\ &= 3x - 2(2x-y) \\ &= -x + 2y. \end{align*}$

Hence, this has the Cartesian equation $x - 2y + z = 0$ . But this is the plane $M'$ ; hence, $M'$ is parallel to $M$ .
The points $(0,0,0)$ and $(-1,0,1)$ are both in the intersection $M' \cap M''$ since they both satisfy the equations
$\begin{align*} x-2y+z &= 0 \\ x + 2y + z&=0. \end{align*}$