Example 11 A convex polyhedron's surface is composed of 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex, one square, one octagon, and one hexagon intersect. Among the line segments connecting the vertices of the polyhedron, how many are inside the polyhedron, not on the faces or edges of the polyhedron?
Solution: Since there are exactly three types of faces meeting at each vertex, and each square's four vertices must be four different vertices of the polyhedron, the number of vertices of the polyhedron is $4 \times$ $12=48$.

Since three edges emanate from each vertex, and each edge is counted twice, the number of edges of the polyhedron is $\frac{1}{2} \times 3 \times 48=72$, and since each square has 2 diagonals, each regular hexagon has $C_{6}^{2}-6=9$ diagonals, and each regular octagon has $C_{8}^{2}$ $-6=20$ diagonals. Therefore, the number of diagonals on the faces of the polyhedron is $12 \times 2+8 \times$ $9+6 \times 20=216$ (diagonals).
Thus, the number of space diagonals of the polyhedron is $C_{88}^{2}-72-216=840$ (diagonals).