5. Given that each edge of a certain cube has three equal division points, and each vertex is cut off by a plane passing through the nearest three equal division points to that vertex, the remaining polyhedron has 24 vertices. How many diagonals (located inside the polyhedron) can be drawn between these vertices?
5. From the problem, we know that the polyhedron has eight triangular faces and six octagonal faces, and each edge is shared by two faces.
Thus, the total number of edges of the polyhedron is
$$
\frac{1}{2}(8 \times 3 + 6 \times 8) = 36.
$$

Since each octagonal face has 20 diagonals, the total number of diagonals on the faces of the polyhedron is $6 \times 20 = 120$.

The number of lines connecting the vertices of the polyhedron is $\mathrm{C}_{24}^{2} = 276$, so the number of internal diagonals of the polyhedron is $276 - 36 - 120 = 120$.