Answer: 30 ways. First of all, note that for each of the 10 large diagonals of the dodecahedron, there are exactly three different planes perpendicular to this diagonal and cutting out a regular hexagon. Indeed, let us move the plane perpendicular to the diagonal from one vertex to another. At first, a regular triangle will be in the section, then an irregular hexagon, which at a certain moment will become regular, then again an irregular hexagon, which will become regular when we reach the center of the dodecahedron; after this, everything will repeat in reverse order. It remains to check that a plane not perpendicular to the large diagonals of the dodecahedron cannot cut out a regular hexagon. For this, the following cases need to be considered: 1) the parallel sides of the regular hexagon lie on two adjacent faces; 2) the parallel sides of the hexagon lie on two non-adjacent faces, bordering the same face; 3) the parallel sides of the hexagon lie on two opposite faces. The first case is impossible. The second case is easily resolved. In the third case, if we take into account the second, it turns out that each pair of parallel sides lies on opposite faces. This case is now also not difficult to resolve. 

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In space, there is a regular dodecahedron. In how many ways can a plane be drawn so that it intersects the dodecahedron to form a regular hexagon?

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