Determine which points are on a line containing (-3,1,1) and (1,2,7)

Let $L$ be the line containing the points $P = (-3,1,1)$ and $Q = (1,2,7)$ . Determine which of the following points are also on $L$ .

$(-7,0,5)$ ;
$(-7,0,-5)$ ;
$(-11,1,11)$ ;
$(-11,-1,11)$ ;
$(-1, \frac{3}{2}, 4)$ ;
$(-\frac{5}{3}, \frac{4}{3}, 3)$ ;
$(-1, \frac{3}{2}, -4)$ .

First, the line containing the points $P = (-3,1,1)$ and $Q = (1,2,7)$ is the set of points

$L(P;P-Q) = \{ (-3,1,1); (-4,-1,-6) \} = \{ (-3-4t, 1-t, 1-6t) \mid t \in \mathbb{R} \}.$

The point $(-7,0,5)$ is not on $L$ since
$-3-4t = -7 \quad \implies \quad t = 1.$

But then $1 - 6t = -5 \neq 5$ . Hence, there is no $t \in \mathbb{R}$ such that $(-7,0,5) = (-3-4t, 1-t, 1-6t)$ , so $(-7,0,5)$ is not on $L$ .
The point $(-7,0,-5)$ is on $L$ since
$-3-4t = -7 \quad \implies \quad t = 1.$

And then $1-t = 0$ and $1-6t = -5$ . Hence, $(-7,0,-5) = (-3-4t,1-t,1-6t)$ where $t = 1$ . Thus, $(-7,0,-5)$ is on $L$ .
The point $(-11,1,11)$ is not on $L$ since
$-3-4t = -11 \quad \implies \quad t = 2.$

But then $1-t = -1 \neq 1$ . Hence, there is no $t \in \mathbb{R}$ such that $(-11,1,11) = (-3-4t, 1-t, 1-6t)$ , so $(-11,1,11)$ is not on $L$ .
The point $(-11,-1,11)$ is not on $L$ since
$-3-4t = -11 \quad \implies \quad t = 2.$

But then $1 - 6t = -11 \neq 11$ . Hence, there is no $t \in \mathbb{R}$ such that $(-11,-1,11) = (-3-4t, 1-t, 1-6t)$ , so $(-11,-1,11)$ is not on $L$ .
The point $(-1,\frac{3}{2},4)$ is on $L$ since
$-3-4t = -1 \quad \implies \quad t = -\frac{1}{2}.$

And then $1-t = \frac{3}{2}$ and $1-6t = 4$ . Hence, $(-1,\frac{3}{2},4) = (-3-4t,1-t,1-6t)$ where $t = -\frac{1}{2}$ . Thus, $(-1,\frac{3}{2},4)$ is on $L$ .
The point $(-\frac{5}{3},\frac{4}{3},3)$ is on $L$ since
$-3-4t = -\frac{5}{3} \quad \implies \quad t = -\frac{1}{3}.$

And then $1-t = \frac{4}{3}$ and $1-6t = 3$ . Hence, $(-\frac{5}{3},\frac{4}{3},3) = (-3-4t,1-t,1-6t)$ where $t = -\frac{1}{3}$ . Thus, $(-\frac{5}{3},\frac{4}{3},3)$ is on $L$ .
The point $(-1,\frac{3}{2},-4)$ is not on $L$ since
$-3-4t = -1 \quad \implies \quad t = -\frac{1}{2}.$

But then $1 - 6t = 4 \neq -4$ . Hence, there is no $t \in \mathbb{R}$ such that $(-1,\frac{3}{2},-4) = (-3-4t, 1-t, 1-6t)$ , so $(-1,\frac{3}{2},-4)$ is not on $L$ .