Question: in the text I am reading, this problem is set before the introduction of points at infinity. Does this mean I need to ignore the case $AA'\parallel BB'$ (which implies $CC'$ is also parallel)? I cannot see a contradiction with the assumption that they are parallel, but without points at infinity it would seem to contravene the theorem.
By assumption we have
for planes $\color{green}\pi$ and $\color{blue}{\pi'}$ not parallel. We wish to show that lines $AA', BB', CC'$ are concurrent.
As a start we note that, since $AB$ and $A'B'$ meet at $Q$, $AA'$ and $BB'$ are in a plane and so must meet at some point $O$. We thus need only show that the line through $CC'$ also passes through $O$.
To see this, consider the planes $\tau$ containing $\triangle$ $ARA'$ and $\tau'$ containing $\triangle CPC'$. These planes are not parallel and so must intersect in a line $l$. The segment $CC'$, being common to both planes, must be on $l$. Further, since the segments $AA'$ and $BB'$ are exclusively in $\tau$ and $\tau'$ respectively, the lines extending from them must meet at a point on $l$, which by above is $O$.
$\square$