Consider the prismatoid with two parallel bases: a pentagon $ABCDE$ at the bottom and a triangle $A_1B_1C_1$ at the top, placed in a way that the orthogonal projection of the triangle on a pentagon's plane will end up inside the pentagon. See the picture for clearer reference.
A point $N$ belongs to plane $ABA_1$, and a point $K$ belongs to a plane $AEB_1$. Using straightedge and compass, build a point $J$ where line $KN$ intersects a plane $BCB_1$.
People I've asked are saying it's commonly not that easy to define such intersection points, but nevertheless I am very curious on how to do it for a generalized and complex case like this one is.
Thanks in advance!
Let's say, in general, you have a plane $PQR$ and a line $AB$ not parallel to $PQR$, and want to find the intersection point $M$ between line and plane.
First of all, we must construct the perpendicular projection $A'$ of $A$ on plane $PQR$. To this end, let $H$ and $K$ be the projections from $A$ to lines $PQ$ and $QR$ respectively (these are standard plane constructions). From $H$ and $K$ draw two lines $HA'$ and $KA'$, on plane $PQR$, perpendicular to $PQ$ and $QR$, meeting at $A'$, which is the desired projection.
Repeat the same procedure to construct the perpendicular projection $B'$ of $B$. The intersection point $M$ is then the intersection point between line $AB$ and line $A'B'$.