§ 17:

The intersection point x.a is a fixed point, so it also lies on x ' .
For any line l we choose through C holds: if X = x. l, then X ' = x ' . l. See O14 for the rest.

Let A be a regular linear transformation of ℜ3 inducing φ, with A(0,0,1) = (0,0,1).
Then from A(1,0,0) = (ρ,0,0), A(0,1,0) = (0,σ,0) and A(1,1,0) = (τ,τ,0) we find ρ = σ = τ.
Furthermore, φ maps the point (1,0,1) to the fourth harmonic with {(0,0,1),(1,0,0)} and (1,0,1); this is (-1,0,1).
So A(1,0,1) = ω(-1,0,1). We also have A(1,0,1) = A(1,0,0) + A(0,0,1) = (ρ,0,1). So ω = 1 and ρ = -1.
Then the matrix becomes ((-1,0,0),(0,-1,0),(0,0,1)), so φ is the reflection in the point (0,0,1) in {x3 = 1}.

Require for three points on λ(1,1,1) they map onto themselves, and require for two points λ(1,1,1) that the image point lies on the line through that point and through λ(0,1,-1). It follows that the matrix has the form ((p,0,0),(q,p+q,q),(-q,-q,p-q)).

Check that the corresponding linear transformations have a twodimensional eigenspace and a onedimensional eigenspace.
Axis λ(-1,9,3), centre λ(-11,-2,1).

Axis λ(1+√2,0,-1), centre λ(1,0,1-√2).
Now check for one point X that {X,X ' } separates the pair {C,SX} harmonically.
Alternative: determine the square of the matrix and show that this is a multiple of the identity matrix.

Start from a linear transformation A with A(0,0,1) = (β,γ,1).
If α=1 and (β,γ) ≠ (0,0), then there is no fixed point outside x3=0.
The line through (0,0,1) and (β,γ,1) intersects x3=0 in the centre (β,γ,0).