The group $PSL(3,4)$ acts on the projective plane $\mathbb{P}_2(\mathbb{F}_4),$ which is the Steiner system $S(2,5,21),$ meaning a collection of pentads (sets of size 5)—the lines of the projective plane—such that each pair of points among the 21 is in exactly one pentad.
After doing some arithmetic, I have not proved it impossible that one could get the Steiner system $S(3,6,22)$ by including one extra point, even with the added condition that those triads that were in a pentad get extended by adding in the extra point—i.e., the sets of three points that were on a line together get the extra point tacked on.
But is it possible? And if so, is there an easily describable way to get the hexad of three noncollinear points? For example, for $(0,0),(1,0),$ and $(0,\omega)$ in the affine part of the plane.
The group $M_{21}\cong PSL(3,4)$ is supposed to be the stabilizer of one point among the 22 points of $S(3,6,22),$ at least as a subgroup of $M_{22},$ which I believe is half the entire symmetry group of the Steiner system.
Another document made me think it could have something to do with hyperconics, which seem like they could be related to conics and thus quadric surfaces (meaning quadratic, but when dealing with homogeneous coordinates and projective spaces). I would have a hard time believing that each set of three noncollinear points is on a unique conic section that happens to include three more points, but perhaps it's true, or if there were an additional restriction on the conic?
I think p. 147 of Robert Peter Hansen (2011) indicates how: "We show that there are 168 hexads in $P_2(\mathbb{F}_4)$, falling into three orbits of 56 under the action of $PSL_3(\mathbb{F}_4)$." Following Witt 1938, he demonstrates that every conic that is regular enough (i.e., not a line and another criterion) has 5 points, and that if one takes the union of the lines connecting those points, there is only one point left out. The union of the conic and that point is called a hyperconic, and I bet that one can take one of the three orbits of 56 mentioned together with the 21 lines of the plane, where each line has the new 22nd point adjoined, to get the 77 desired sets.
Some Sage code I wrote revealed that there are 1029 conics with 5 points, 210 with 9 points, and 21 with 1 point. Subtracting out the 21 lines, there should be 1008 5-point conics that can be extended to 6-point hyperconics. It makes sense that each hyperconic corresponds to 6 conics by removal of an arbitrary point.
I finished the code and it avers that the said 77 sets of 6 points in the projective plane with an extra point added do indeed make a Steiner system S(3,6,22).