Let's find the points on the surface of the cube that are equidistant from the two endpoints of any body diagonal.
The points in question are cut out from the cube by the perpendicular bisector plane of the body diagonal. This plane contains the center of the cube and bisects all edges of the cube that do not originate from either endpoint of the selected body diagonal. The plane cuts out a half-diagonal segment of length $\frac{a \sqrt{2}}{2}$ from each face of the cube, thus forming a regular hexagon in the cube.

![](https://cdn.mathpix.com/cropped/2024_05_02_59fb9d50ab45750048adg-1.jpg?height=307&width=312&top_left_y=318&top_left_x=901)

Every vertex of the hexagon is equidistant, specifically $\frac{a \sqrt{2}}{2}$ away from the center of the cube, so the hexagon is not only equilateral but also regular.