Example 6 Given four parallel planes $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ arranged in sequence, where the distance between each adjacent pair of planes is 1. If the four vertices of a regular tetrahedron $A_{1} A_{2} A_{3} A_{4}$ satisfy $A_{i} \in \alpha_{i}(i=1,2,3,4)$, find the volume of the regular tetrahedron.
【Analysis】Since the distance between two planes can be converted into the distance between a point and a plane, we can try to place the regular tetrahedron inside a cube by considering the relationship between a regular tetrahedron and a cube, and then imagine the four planes.

Solution As shown in Figure 7, place the regular tetrahedron \(A_{1} A_{2} A_{3} A_{4}\) inside the cube \(A B C D-A_{1} B_{1} C_{1} D_{1}\) (or by adding a right triangular pyramid outside each of the four faces of the tetrahedron to form a cube), where \(E_{1}\) and \(F_{1}\) are the midpoints of edges \(A_{1} B_{1}\) and \(C_{1} D_{1}\) respectively, and plane \(E E_{1} D_{1} D \parallel\) plane \(B B_{1} F_{1} F\). If the distance between the two planes is 1, then the tetrahedron \(A_{1} A_{2} A_{3} A_{4}\) is the regular tetrahedron that meets the condition.

Figure 8 is the top face of the cube, with the edge length of the cube set as \(a\).
If \(A_{1} M = M N = 1\), then
\[
A_{1} E_{1} = \frac{a}{2}
\]
\[
D_{1} E_{1} = \sqrt{A_{1} D_{1}^{2} + A_{1} E_{1}^{2}} = \frac{\sqrt{5}}{2} a
\]
By \(A_{1} D_{1} \cdot A_{1} E_{1} = A_{1} M \cdot D_{1} E_{1}\)
\[
\Rightarrow a = \sqrt{5}
\]
\[
\Rightarrow \text{the edge length of the tetrahedron } d = \sqrt{2} a = \sqrt{10},
\]
its volume \(V = a^{3} - 4 \times \frac{1}{6} a^{3} = \frac{1}{3} a^{3} = \frac{5 \sqrt{5}}{3}\),
indicating that \(V\) is equal to the volume of a cube with edge length \(a\) minus the volumes of four right triangular pyramids.