Let $ T=ABCD \in \mathbb{RP}^3$ be a tetrahedron. Let $l$ be a line through $A$ not contained in any of the tetrahedrons faces. Let $ f: \mathbb{RP}^3 \to \mathbb{RP}^3, B \mapsto C, C \mapsto D, D \mapsto B, \forall x \in l: x \mapsto x$ be a projective transformation fixing the points on line $l$. Show that its only fix points are on $l$. Determine the invariant planes of $f$.

I already showed that $f$ is unique by taking a second point $P \in l$ on the line not contained in any of the tetrahedrons faces. Thus $A, B, C, D, P$ lie in general position and determine the unique projectivity $f$. I tried fixing $A = (0:0:0:1), B = (1:0:0:0), C = (0:1:0:0), D = (0:0:1:0)$, thus $l ={ax+by+cz=0}$, and choosing $P = (bc:ac:-2ab:bc)$ to show that the linear map associated to $f$ has only two linearly independent eigenvectors which are represented by $A$ and $P$ thus spanning the line $l$. But i failed along the way and also suspect there might be a quicker way of showing this. For the invariant plane i suspect $BCD$ is one, but am unsure how to show that.

Help appreciated!


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