Let $L$ be the line in $\mathbb R^3$ that fits through the points $P = (1, 2, 3)$ and $Q = (-1, 1, -1)$. Determine a parametric form of $L$.
So I'm assuming that the task is to determine a scalar $t$ and a vector $v$ so that I can express the line as such:
$(x, y, z,) = (x_{0}, y_{0}, z_{0}) + t \times v$
Where $(x, y, z)$ is $P$ and $(x_{0}, y_{0}, z_{0})$ is $Q$ or vice versa. But I'm unsure of how I would go about doing this.
You are mostly correct, but there's no need to determine $t$: we'll be leaving that free as our parameter (also, your $(x,y,z)$ will be an arbitrary point on the line, rather than just $P$).
Thus, what we need is a vector going from $P$ in the direction of $Q$ (or vice versa). Shifting everything by $P$ (that is: subtracting $P$ from everything), we want a vector from the origin in the direction of $Q - P$. Well, there's a very obvious such vector: it's $Q - P$ itself.
Now, the parametric equation $(x(t),y(t),z(t)) = t(Q - P)$ doesn't quite work: it's in the right direction, but it goes through the origin, instead of wherever our line is. To fix that, we just need to pick any point on our line to shift it by: to keep things simple, we'll pick $P$. Thus, our parametric equation is exactly
$$(x(t),y(t),z(t)) = P + t(Q - P).$$
I'll leave you to substitute in your particular values for yourself.