Determine which points are on a plane given by a parametric equation

Let $M$ be a plane defined by the scalar parametric equations

$x = 1+s-2t, \qquad y = 2+s+4t, \qquad z = 2s+t.$

Determine which of the following points are on $M: \ \ (0,0,0), \ (1,2,0), \ (2,-3,-3)$ .
Find vectors $P,A,B$ such that $M = \{ P + sA + tB \}$ .

The point $(0,0,0)$ is not on $M$ since
$2s+t = 0 \quad \implies \quad t = -2s.$

Then,

$2+s+4t = 0 \quad \implies \quad 2+s+4(-2s) = 0 \quad \implies \quad s = \frac{2}{7}, \quad t = -\frac{4}{7}.$

But then the first coordinate fails since

$1+s-2t = 1 + \frac{2}{7} + \frac{8}{7} = \frac{17}{7} \neq 0.$

The point $(1,2,0)$ is on $M$ since the system of equations

$\begin{align*} 1+s-2t &= 1 \\ 2 + s + 4t &= 2 \\ 2s + t &= 0 \end{align*}$

has a solution $s = t = 0$ . Therefore, $(1,2,0) = (1+s-2t, 2+s+4t,2s+t)$ for $s = t = 0$ .

The point $(2,-3,-3)$ is on $M$ since the system of equations

$\begin{align*} 1+s-2t &= 2 \\ 2+s+4t &= -3 \\ 2s+t &= -3 \end{align*}$

has a solution $s = t = -1$ .
Since $(1,2,0)$ is on the plane we take $P = (1,2,0)$ . Then from the parametric equations we get the vectors $A$ and $B$ as the coordinates of $s$ and $t$ . So, $A = (1,1,2)$ and $B = (-2,4,1)$ . Then we have
$M = \{ (1,2,0) + s(1,1,2) + t(-2,4,1) \}.$