TITLE: Seeking a more symmetric realization of a configuration of 10 planes, 25 lines and 15 points in projective space
QUESTION [15 upvotes]: 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.

REPLY [13 votes]: With the aid of the answer I now managed to find a better realization. At the expense of some of the symmetries it can be nicely drawn in 3d space - it is just the barycentric subdivision of a tetrahedron:

The planes are faces (4 of them) and the planes through an edge and the barycenter of the tetrahedron (+6); the lines are edges (6), the lines joining a vertex with the barycenter of its opposite face (+4) or with the midpoint of one of its non-incident edges (+12), as well as lines through midpoints of opposite edges (+3); and the points are vertices, edge midpoints, face barycenters, and the barycenter of the tetrahedron (4+6+4+1). Additional symmetry that is not well apparent in this realization is that there is a full S$_5$. In the projective 3-space it acts transitively on all five yellow points (i. e. vertices of the tetrahedron and its barycenter), on all ten blue points, all ten orange lines, all fifteen green lines, and all ten planes.
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