# Seeking a more symmetric realization of a configuration of 10 planes, 25 lines and 15 points in projective space

I've got ten (projective) planes in projective 3-space:

\begin{align} &x=0\\ &z=0\\ &t=0\\ &x+y=0\\ &x-y=0\\ &z+t=0\\ &x-y-z=0\\ &x+y+z=0\\ &x-y+t=0\\ &x+y-t=0 \end{align} If I did not make a mistake somewhere, their intersections produce $$25$$ lines and $$15$$ points, each line containing $$3$$ points, each plane containing $$6$$ lines and $$7$$ points. Ten of the points belong to $$4$$ lines and $$4$$ planes each and five of them belong to $$7$$ lines and $$6$$ planes each. $$15$$ lines belong to $$2$$ of the planes and $$10$$ of them belong to $$3$$ of the planes.

The above indicates that this configuration is highly symmetric, is it known? How to compute its automorphism group? Where to look? My goal is to find another more symmetric realization.

This configuration has automorphisms by the symmetric group $$S_5$$, and can be identified with the planes $$a_i = a_j$$ ($$0 \leq i < j \leq 4$$) in the projective 3-space $$a_0+a_1+a_2+a_3+a_4 = 0$$, by setting $$(x,z,t,y) = (a_0 - a_1, 2(a_2 - a_3), 2(a_4 - a_2), a_0 + a_1 - 2a_2).$$ the ten planes then have $$(i,j)$$ in the order $$(0,1),\,(2,3),\,(2,4),\,(0,2),\,(1,2),\,(3,4),\,(1,3),\,(0,3),\,(1,4),\,(0,4).$$ The 25 lines include the ten lines $$a_i=a_j=a_k$$ where three of the planes $$a_i=a_j$$ meet; these ten lines together with the ten planes $$a_i = a_j$$ encode Desargues configurations: intersecting with a generic plane $$\Pi$$ in the 3-space yields the ten points and ten lines of a Desargues configuration, and any Desargues configuration can be obtained this way for some choice of $$\Pi$$.