COURSE OF PROJECTIVE GEOMETRY
Find a 2 by 2 matrix whose determinant is to be 0.
We get x' y' = z' (ρ z' - τ x').
By substitution of the coordinates of the five points in the equation we get five linear equations with six unknown variables aij. In general, this gives a 1-dimensional solution space, so the coefficients aij are then unique up to a scalar factor.
(These five linear equations are linearly independent if and only if the five points are freely situated.)