The question is find the vector equation for the line passing through two points P1 = (6,2,4) and P2 = (12,0,3) But the formatting of the answer is throwing me off. Should I just find the difference between these two points?
The answer format is as follows: [x,y,z] = [0,0,0] + t[0,0,0]
The equation of the line in $\mathbb{R}^n$ can be represented as $\vec{x} = \vec{u} + t \vec{v}$, where $\vec{u}, \vec{v} \in \mathbb{R}^n$ are some constants and $\vec{x} = (x_1, x_2, \ldots, x_n)$ are variables. This form captures a linear relationship between the variables, i.e. in each unit of $t$, all variables grow in the same proportion, governed by $\vec{v}$, and $\vec{u}$ is used to translate the line passing through the origin to allow it to pass through any point in $\mathbb{R}^n$.
In your case, 2 points are given, $P = (6,2,4)$ and $Q = (12, 0, 3)$. How fast $t$ changes is not important, since it does not alter the relative rates of change of the individual variables. So $\vec{v}$ measuring change will be given by $$\vec{v} = \vec{Q} - \vec{P} = (6, -2, -1)$$ and now the line looks like $$(x,y,z) = \vec{u} + t (6,-2,-1).$$
You need to pick $\vec{u}$ so that the line passes through both $P$ and $Q$. If we make it pass through one of them, it will automatically pass through the other one since that was the whole idea of fixing $\vec{v}$.
So let's make it pass through $\vec{P}$. To do that, note that when $t = 0$, we have $(x,y,z) = \vec{u}$, i.e., the line always passes through $\vec{u}$, no matter which values you pick for it. So letting $\vec{u} = \vec{P}$ we end up with the general equation $$ \vec{x} = \vec{P} + t \left( \vec{Q} - \vec{P} \right), $$ which in our case yields $$ (x,y,z) = (6,2,4) + t (6, -2, -1) $$ which is equivalent to simultaneous equations $$ \begin{cases} x = 6+6t\\ y = 2-2t\\ z = 4-t \end{cases} $$