jmc

Every pair of vertices of the polyhedron determines either an edge, a face diagonal or a space diagonal. We have $\left(\genfrac{}{}{0ex}{}{26}{2}\right)=\frac{26\cdot 25}{2}=325$ total line segments determined by the vertices. Of these, $60$ are edges. Each triangular face has $0$ face diagonals and each quadrilateral face has $2$, so there are $2\cdot 12=24$ face diagonals. This leaves $325-60-24=\boxed{241}$ segments to be the space diagonals.