Determine which points are on a given plane

Let

$P = (1,2,-3, \qquad A = (3,2,1), \qquad B = (1,0,4).$

Then, define the plane $M = \{ P + sA + tB \}$ . Determine which of the following points are on $M$ .

$(1,2,0)$ ;
$(1,2,1)$ ;
$(6,4,6)$ ;
$(6,6,6)$ ;
$(6,6,-5)$ .

First, we have

$M = \{ P + sA + tB \} = \{ (1,2,-3) + (3s,2s,s) + (t,0,4t) \} = \{ (1+3s+t, 2+2s, -3+s+4t) \}.$

The point $(1,2,0)$ is not on $M$ since
$2+2s = 2 \quad \implies \quad s = 0$

and

$1+3s + t = 1 \quad \implies \quad t = 0.$

But then, $-3 + s + 4t = -3 \neq 0$ .
The point $(1,2,1)$ is not on $M$ since
$2+2s = 2 \quad \implies \quad s = 0$

and

$1+3s + t = 1 \quad \implies \quad t = 0.$

But then, $-3 + s + 4t = -3 \neq 1$ .
The point $(6,4,6)$ is on $M$ since
$2+2s = 4 \quad \implies \quad s = 1$

and

$1+3s+t = 6 \quad \implies \quad t = 2.$

Then, $-3+s+4t = 6$ . Hence, $(6,4,6) = (1+3s+t, 2+2s, -3+s+4t)$ for $s = 1, \ t = 2$ .
The point $(6,6,6)$ is not on $M$ since
$2+2s = 6 \quad \implies \quad s = 2$

and

$1+3s + t = 6 \quad \implies \quad t = -1.$

But then, $-3 + s + 4t = -5 \neq 6$ .
From part (d) we immediately see that $(6,6,-5)$ is on $M$ .