Let $A$ and $G$ be two opposite vertices of a unit cube. What is the distance between the plane $S_{A}$ determined by the vertices adjacent to $A$, and the plane $S_{G}$ determined by the vertices adjacent to $G$?
I. Solution. We will show that the planes $S_{A}$ and $S_{G}$ trisect the body diagonal $A G$ of the cube, and thus their distance is $\frac{A G}{3}=\sqrt{3} / 3$.

The $120^{\circ}$ rotations around the $A G$ diagonal map the cube onto itself. Therefore, during these rotations, the planes $S_{A}$ and $S_{G}$ map onto themselves, meaning both planes are perpendicular to the body diagonal $A G$. Due to symmetry, the distance from point $G$ to the plane $S_{G}$ is equal to the distance from point $A$ to the plane $S_{A}$. Therefore, it is sufficient to show that, for example, the latter distance is equal to the distance between the planes $S_{A}$ and $S_{G}$.

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Figure 1

Reflect the cube over the midpoint of the face diagonal $B D$. Thus, we obtain a cube $A B C D G^{\prime} H^{\prime} E^{\prime} F^{\prime}$ that shares a face with the original one (Figure 1). The reflection of the plane $S_{A}=E B D$ is itself, and thus the distance from point $A$ to the plane $S_{A}$ is equal to the distance from the reflected point and the reflected plane, i.e., the distance from point $C$ to the plane $S_{A}$. The latter is precisely the distance between the planes $S_{A}$ and $S_{G}$, since point $C$ lies in the plane $S_{G}$. This completes the proof.

II. Solution. Label the vertices of the cube as shown in Figure 1, and consider the vectors $\overrightarrow{A B}, \overrightarrow{A D}$, and $\overrightarrow{A E}$ from point $A$ to the adjacent vertices (Figure 2).

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Figure 2

It is known that the vector from point $A$ to the centroid $T_{A}$ of triangle $E B D$ is the arithmetic mean of the vectors to the vertices of the triangle, $\overrightarrow{A T_{A}}=\frac{1}{3}(\overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{A E})$. On the other hand, clearly $\overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{A E}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C G}=\overrightarrow{A G}$, so $\overrightarrow{A G}=3 \overrightarrow{A T_{A}}$.

The obtained equality means that points $A, T_{A}$, and $G$ lie on the same line, and the point $T_{A}$, and thus the plane $S_{G}$, trisects the segment $A G$. This is of course also true for the plane $S_{G}$. Since the body diagonal $A G$ is perpendicular to the planes $S_{A}$ and $S_{G}$, the distance between the two planes is one-third of the body diagonal, $\sqrt{3} / 3$.

Remark. In both solutions, only the perpendicularity of the planes in question and the body diagonal to the cube's special properties (3rd order rotational symmetry) was needed for the proof, so the solutions generally show that in any parallelepiped, the planes through the vertices adjacent to the endpoints of a body diagonal trisect the body diagonal.