§ 17:

O65
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.

O66
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}.

O67
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)).

O68
Check that the corresponding linear transformations have a twodimensional eigenspace and a onedimensional eigenspace.
Axis λ(-1,9,3), centre λ(-11,-2,1).

O69
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,S_X} harmonically.
Alternative: determine the square of the matrix and show that this is a multiple of the identity matrix.

O70
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).