Math Kangaroo Problem 3D Geometry "Question 24
How many planes pass through exactly three vertices of a given cube?
(A) 1 (B) 2 (C) 4 (D) 8 (E) 12"
This was one only problems that did not make sense to me on the test although it seems so easy. The reason I do not understand it is that a plane can go through 1 vertex (tangent), it can go through no vertices, it can go through 2 vertices (tangent to a edge), and it can go through 4 vertices by being on or diagonal to one of the sides. How can a plane go through 3 vertices? By plane does it have to be an infinitely endless rectangular plane? Any help through diagrams or answers would be helpful!
 A: The only way a plane can go through exactly $3$ vertices in a cube is through $3$ face diagonals that form a triangle (see the diagram).
The moment a plane includes an edge, it either goes through $2$ vertices or through $4$ vertices. Same is true for a plane going through a space diagonal.

Now the counting:
i) Start with one of the vertices, say vertex $B$. There are $3$ face diagonals through $B$. There are $3$ ways to choose two face diagonals that will uniquely identify a plane.
ii) Then vertex $A$. As $AB$ is an edge, we had no plane counted in (i) that included vertex $A$. So we again have $3$ desired planes through vertex $A$.
iii) Out of $3$ planes through vertex $C$, two of them include diagonal $AC$ so those two are duplicates as they were counted in $(ii)$.
iv) Same with planes through vertex $D$, two of them include diagonal $BD$ so two duplicates, already counted in $(i)$.
Please note that we do not have to count planes through vertices $E, F, G, H$ meeting the criteria as any plane through these vertices meeting the criteria would involve a vertex from the bottom face $ABCD$ which we already counted.
So total number of planes going through exactly $3$ vertices,
$ = 3 + 3 + (3-2) + (3-2) = 8$
