Is it possible to find $6$ points $a_0,a_1,\dots,a_5$ such that the tetrahedra $T_i$ defined by $\{a_i,a_{i+1},a_{i+2},a_{i+3}\}$ (defined for $i\in\{0,1,2,5\}$ using $\bmod 6$) satisfy that the pairwise intersections of their convex hulls are the convex hulls of their intersections?
I have tried a bunch of things but nothing seems to work.
Unless I've made a miscalculation, these points ought to work:
$$ a_0 = (0, 2,0) \\ a_1 = (0, 2, 4) \\ a_2 = (-1,0,-2) \\ a_3 = (-2,0,-1) \\ a_4 = (4,-2,0) \\ a_5 = (0,-2,0) $$
The plane containing $a_5a_0a_1$ has equation $x=0$, and $a_4$ and $a_2$ are on opposite sides of it.
The plane containing $a_0a_1a_2$ has equation $2x-y = -2$, and $a_5$ and $a_3$ are on opposite sides of it.
The plane containing $a_1a_2a_3$ has equation $2x-7y+2z = -6$, and $a_0$ and $a_4$ are on opposite sides of it.
The remaining $3$ planes work similarly, by the symmetry $$ a_i = (x,y,z) \mapsto a_{5-i} = (z,-y,x) $$