# Linearly dependent planes

Let $$A$$ is set of all possible planes passing through four vertices of given cube. Find number of ways of selecting four planes from set $$A$$, which are linearly dependent and have one common point.

I know that if planes $$P_1=0,P_2=0,P_3=0$$ and $$P_4=0$$ can be written as $$aP_1+bP_2+cP_3+dP_4=0$$, where all $$a,b,c$$ are not equal to zero, then we can say that $$P_1,P_2,P_3,P_4$$ are linearly dependent planes.

How to proceed?

$$135$$.

There are in total $$12$$ planes in $$A$$: six for the faces, and six through the cube's center. But the answer is not $$\binom{12}{4}$$.

• First of all, are you able to denumerate how many planes are passing through 4 vertices of a cube ? Jul 23, 2020 at 14:49

First of all, $$n(A)=12$$ (6 faces and 6 planes through face diagonals).
Every vertex of the cube has 6 planes passing through it. We can Select 4 planes from these 6 in $$\binom 64 = 15$$ ways. Since there are 8 vertices, the number of sets of 4 planes intersecting at the vertices is $$15 \times 8 = 120$$.
The geometric center of the cube also has 6 planes passing through it (the 6 planes along the face diagonals). Thus, there are 15 sets of 4 planes intersecting at this point. This brings the total to $$120+15 = \boxed{135}$$.
For a more exhaustive proof, you can also prove that there are no other points in $$R^3$$ where more than 3 such planes intersect.
• [+1] Very good solution. But the asker hasn't done really any work (unable for example to answer my question about $n(A)$). Jul 23, 2020 at 21:39