380 The four planes in space are mutually parallel, and the distance between each adjacent pair of planes is $h$. A regular tetrahedron has its vertices on the four planes. Find the edge length of the regular tetrahedron.
Solve As shown in Figure 2, the vertices of the regular tetrahedron $ABCD$ lie on four parallel planes $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$.

Let $AD$ intersect planes $\alpha_{2}, \alpha_{3}$ at points $E, F$ respectively, and $AC$ intersect plane $\alpha_{2}$ at point $H$. Connect $BE, EH, HB$.

For convenience, let $AB=6x$.
From the given, it is easy to see that
$$
AE=EF=FD=2x, AH=HC=3x.
$$

Applying the cosine rule to $\triangle ABE, \triangle AEH, \triangle AHB$ yields
$$
BE=2\sqrt{7}x, EH=\sqrt{7}x, HB=3\sqrt{3}x.
$$

In $\triangle BEH$, by the cosine rule we get
$$
\begin{array}{l}
\cos \angle BHE=\frac{HB^2+EH^2-BE^2}{2HB \cdot EH}=\frac{1}{\sqrt{21}} \\
\Rightarrow \sin \angle BHE=\frac{2\sqrt{5}}{\sqrt{21}} \\
\Rightarrow S_{\triangle BEH}=\frac{1}{2}HB \cdot EH \sin \angle BHE=3\sqrt{5}x^2. \\
\text{Also, } V_{\text{tetrahedron } BHEH}=\frac{1}{6}V_{\text{tetrahedron } 4ABCD}=3\sqrt{2}x^3=\frac{h}{3}S_{\triangle BEH} \\
\Rightarrow 6x=\sqrt{10}h.
\end{array}
$$

Therefore, the edge length of the regular tetrahedron $ABCD$ is $\sqrt{10}h$.
(Huang Quanfu, Jiangzhen Middle School, Huaining County, Anhui Province, 246142)