Can you embed a non-degenerate Pappus configuration (that is, all 9 points are distinct) in the projective plane of order 3?
I've been trying to create a nice graphic which showcases a non-degenerate Pappus configuration in a finite projective plane. Clearly this is impossible in the Fano plane (not enough points) so I decided to try the projective plane of order 3 (which has 14 points) but after an hour of trying I haven't managed to do so. I was wondering if this is in fact impossible (and if so, why?)
In the projective plane over the field with three elements, choose the collinear triples of points $A[0,0,1]$, $B[0,1,1]$, $C[0,2,1]$ and $A'[2,1,0]$, $B'[1,0,0]$, $C'[1,1,0]$. (The first three points are on the line $x_1=0$ and the elements of the second triple are on $x_3=0$.) Then an easy calculation shows that the remaining three points of the Pappus configuration are $$AB'\cap A'B=[1,0,1],$$ $$AC'\cap A'C=[1,1,1],$$ $$BC'\cap B'C=[1,2,1].$$ They are in fact collinear, on the line $x_1+2x_3=0$.