Question 10

A tetrahedron has 4 vertices (say $A, B, C, D$) and 6 edges connecting pairs of vertices ($AB, AC, AD, BC, BD, CD$).

Let $P$ be a point strictly inside the tetrahedron. Straight lines are drawn connecting $P$ to the 4 corners: $PA, PB, PC, PD$.

A plane is uniquely determined by 3 non-collinear points. A new internal plane formed by these connecting lines must contain the point $P$ and two vertices of the tetrahedron.

Every pair of vertices $\{V_i, V_j\}$ forms an edge of the tetrahedron. The plane passing through $P$ and that edge (points $P, V_i, V_j$) cuts through the interior of the tetrahedron.

The number of pairs of vertices is:

Each of these 6 planes ($PAB, PAC, PAD, PBC, PBD, PCD$) is a distinct internal plane cutting through the interior of the tetrahedron.

Hence, exactly 6 new internal planes are created.