Example 3 Four planes in space are mutually parallel, and the distance between adjacent planes is $h$. Now there is a regular tetrahedron, whose four vertices are on these four planes respectively. Find the edge length of the regular tetrahedron.
Figure $21-2$
Solution Let the circumscribed cube of the regular tetrahedron $A B C D$ be $A C^{\prime} B D^{\prime}-A^{\prime} C B^{\prime} D$. Suppose the section through the midpoints $F$ of edges $D^{\prime} D$ and $B^{\prime} C$ is $\alpha_{3}$, the section through the midpoints $E$ of edges $C^{\prime} C$ and $A^{\prime} D$ is $\alpha_{2}$, and the planes through edges $A^{\prime} A$ and $B^{\prime} B$ parallel to $\alpha_{3}$ and $\alpha_{2}$ are $\alpha_{1}$ and $\alpha_{4}$, respectively. These four planes are mutually parallel with equal distances between adjacent planes. Suppose the line $l$ through the midpoint $O^{\prime}$ of $A^{\prime} B^{\prime}$ and perpendicular to $C E$ intersects $\alpha_{4}$, $\alpha_{3}$, $\alpha_{2}$, and $\alpha_{1}$ at points $B^{\prime \prime}$, $D^{\prime \prime}$, $C^{\prime \prime}$, and $A^{\prime \prime}$, respectively, as shown in Figure $21-2(\mathrm{~b})$. Then the distances between adjacent planes $\alpha_{4}$, $\alpha_{3}$, $\alpha_{2}$, and $\alpha_{1}$ are $B^{\prime \prime} D^{\prime \prime}$, $D^{\prime \prime} C^{\prime \prime}$, and $C^{\prime \prime} A^{\prime \prime}$.

When $A^{\prime \prime} C^{\prime \prime}=h$, we can find that $A^{\prime} E=\frac{\sqrt{5}}{2} h$, and thus $A^{\prime} B^{\prime}=\sqrt{10} h$. This is the edge length of the regular tetrahedron we are looking for.