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?
Answer given is
$135$.
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, $n(A)=12$ (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.