If I have three collinear points in the $x$-$y$ plane, $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, and each has an associated point in the $z$ direction, how do I derive the equation for a parabola that passes through each of the $z$ points?
Also, is there any way to prove that if the parabola was projected on to the $x$-$z$ or $y$-$z$ planes, that the resulting curve would also be a parabola (I have a feeling it would be)?
It is not clear from the question what kind of equation is expected. If $y=mx+q$ is the equation of the line in the $x-y$ plane, I think the most natural way to handle such a parabola is that of finding the equation of its projection on the $x-z$ plane, which is a parabola of equation $$ \tag{1} z=ax^2+bx+c. $$ You can find $a$, $b$ and $c$, as usual, by plugging there the coordinates of the three given points: only $x$ and $z$ are needed, because $y=mx+q$.
In other words: the parabola is the intersection between the parabolic cylinder given by equation $(1)$ and the plane of equation $y=mx+q$.
If the line in the $x-y$ plane is parallel to the $y$ axis then you can instead project the parabola onto the $y-z$ plane.