Determine if a given set of points all lie on the same line

In each of the following cases determine if the given set of points all lie on the same line.

The points $P,Q,R$ are not on the same line since the unique line containing $P,R$ is given by
$L = \{ P + t(P-R) \mid t \in \mathbb{R} \} = \{ (2,1,1) + t(-1,2,0) \mid t \in \mathbb{R} \} = \{ (2-t,1+2t,1) \mid t \in \mathbb{R} \}.$

But, $Q$ cannot be on this line since for any $t$ , the third coordinate of every point on the line will be 1, while the third coordinate of $Q$ is $-1$ .
The points $P,Q,R$ are not on the same line since the unique line containing the points $P,Q$ is given by
$L = \{ P + t(P-Q) \mid t \in \mathbb{R} \} = \{ (2,2,3) + t (4,-1,2) \mid t \in \mathbb{R} \} = \{ (2+4t, 2-t, 3+2t) \mid t \in \mathbb{R} \}.$

But, $R$ is not on this line since if $-6 = 2+4t$ then we must have $t = -2$ . However, with $t = -2$ we have $3+2t = -1 \neq 1$ .
The points $P,Q,R$ are not on the same line since the unique line containing the points $P,Q$ is given by
$L = \{ P + t(P-Q) \mid t \in \mathbb{R} \} = \{ (2,1,1) + t(4,-2,0) \mid t \in \mathbb{R} \} = \{ (2+4t,1-2t,1) \mid t \in \mathbb{R} \}.$

But then $R$ is not on this line since if $2+4t = 5$ then $t = \frac{3}{4}$ , but then $1-2t = -\frac{1}{2} \neq -1$ . Hence, $R$ is not on the line.