How many planes are there that are perpendicular to the line connecting two opposite vertices of a regular dodecahedron with edge length $a$, and intersect the dodecahedron in a regular hexagon? What are the side lengths of these hexagons?
Consider the $H L$ axis of the dodecahedron as shown in Figure 1, and use the additional notations from the figure.

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Figure 1

Move the plane $S$, which is perpendicular to $H L$ and passes through $H$, from $H$ to $L$. Since the dodecahedron rotates $120^{\circ}$ around $H L$ and maps onto itself, there will be a position of $S$ where $E, F$, and $G$ lie on it. Up to this point, the plane section is a triangle. As $S$ continues to move, we obtain hexagonal sections, analogous to the previous case. First, consider the case where $S$ intersects the edges $E A$ and $E D$ at some internal points. Let the two adjacent edges of the section be $c_{1}$ and $c_{2}$. Since the dodecahedron rotates $120^{\circ}$ and maps onto itself, the section is regular if and only if $c_{1}=c_{2}$. In Figure 2, $c_{1}$ and $c_{2}$ are marked on the same pentagonal face.

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Figure 2

It is known that the diagonal of a regular pentagon with side length $a$ is $\frac{a}{2}(1+\sqrt{5})$. If the ratio $\frac{E D^{\prime}}{E D}$ is denoted by $k$, then $c_{1}=k \cdot \frac{a}{2}(1+\sqrt{5})$. On the other hand, from the trapezoid $B C D E$, we have $B E - c_{2} = k(B E - a)$, which gives $c_{2} = \frac{a}{2}(1+\sqrt{5}) - k \cdot \frac{a}{2}(-1+\sqrt{5})$.

From the condition $c_{1}=c_{2}$, we get $k=\frac{5+\sqrt{5}}{10}$, and in this case, $c_{1}=c_{2}=\frac{a}{10}(5+3 \sqrt{5})$.

If the plane $S$ passes through the point $A$, the section is not regular. As $S$ continues to move, the side $b_{1}$ parallel to $A D$ decreases, while $b_{2}$ increases. The section will be regular precisely when $b_{1}=b_{2}$, which occurs when the cutting plane passes through the center of the dodecahedron, $O$. In this case, due to the central symmetry of the dodecahedron, $S$ passes through the midpoints of the edges it intersects. The side of the hexagon can then be calculated as the midline of the trapezoid $A B C D$:

$$
b_{1}=b_{2}=\frac{1}{2}\left(a+\frac{a}{2}(1+\sqrt{5})\right)=\frac{a}{4}(3+\sqrt{5})
$$

As $S$ continues to move, we find another regular hexagonal section similar to the first case. Therefore, there are 3 planes that are perpendicular to $H L$ and intersect the dodecahedron in a regular hexagon. Since any such plane corresponds uniquely to the $H L$ axis, and the 20 vertices of the dodecahedron determine 10 axes, the number of such planes is 30.