Given a line $l$ through a tetrahedron $ABCD$ (not intersecting any of its edges), take the four points $P_1, P_2, P_3, P_4$ of intersection of the line with the faces of the tetrahedron. Also take the four planes $π_1, π_2, π_3, π_4$ determined by the line and the vertices of the tetrahedron respectively.
Show that the cross ratio of the four points is the same as the cross ratio of the four planes (when choosing a particular ordering), so $$ (P_1, P_2, P_3, P_4) = (π_1, π_2, π_3, π_4) $$
I think i already have an intuitive understanding of why this is true. Each vertice of the tetrahedron has exactly one opposite face. So i would assume as for the particular ordering the plane $π_1$ is that one that goes through the vertice that is opposite of the face which includes $P_1$ and so on. Now i think there could be a projective transformation somehow mapping the planes to the points, thus preserving the cross ratio. Unfortunately i have no idea how to show this actually exist. Was wondering if one could just take clever coordinate representatives for the points and show this using linear algebra, but am also stuck with the particular choice. Appreciate any help!