1.68 A convex polyhedron has a surface composed of 12 squares, 8 regular hexagons, and 6 regular octagons, and each vertex is the common vertex of one square, one octagon, and one hexagon. Among the line segments connecting the vertices of the polyhedron, how many are inside the polyhedron, and not on the faces or edges of the polyhedron?
[Solution] Let $a, b, c, d$ represent the number of vertices, edges, face diagonals, and space diagonals of the convex polyhedron, respectively.

Since there are exactly 3 different faces meeting at each vertex, the vertices of different squares must be different vertices of the polyhedron, so $a=4 \times 12=48$. Since each vertex is connected to 3 edges and each edge connects two vertices, we have $b=\frac{3 a}{2}=72$.

Each square face has 2 face diagonals, so 12 square faces have 24 face diagonals in total. Each hexagonal face has $C_{6}^{2}-6=9$ face diagonals, so 8 hexagonal faces have 72 face diagonals in total. Each octagonal face has $C_{8}^{2}-8=20$ face diagonals, so 6 octagonal faces have 120 face diagonals in total. Therefore, the convex polyhedron has $c=216$ face diagonals.
Since the polyhedron is convex,
$$
d=C_{48}^{2}-b-c=1128-72-216=840,
$$

which means the convex polyhedron has 840 space diagonals.