Draw an image of a cube obtained by orthographic projection of the cube onto a plane perpendicular to the cube's diagonal

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Consider the orthogonal projection of the cube $A B C D A 1 B 1 C 1 D 1$ (Fig.1) with edge $a$ onto the plane $\gamma$, perpendicular to the diagonal $B D 1$ and passing through the vertex $D 1$. In this projection, the vertices $B$ and $D 1$ will coincide at point $D 1$. Let $\alpha$ be the angle between the edge $D D 1$ and the diagonal $B D 1$ (Fig.2). Then the angle between the line $D D 1$ and the plane $\gamma$ is $90^{\circ}-\alpha$. From the right triangle $B D D 1$, we find that

$$
\cos \alpha=\frac{D D_{1}}{B D_{1}}=\frac{a}{a \sqrt{3}}=\frac{1}{\sqrt{3}}, \sin \alpha=\frac{B D}{B D_{1}}=\frac{a \sqrt{2}}{a \sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}} \text {. }
$$

Thus, the orthogonal projection of the edge $D D 1$ onto the plane $\gamma$ is

$$
D D 1 \cos \left(90^{\circ}-\alpha\right)=a \sin \alpha=\frac{a \sqrt{2}}{\sqrt{3}}
$$

Since the edges $A A 1, C C 1$ and $B B 1$ are equal and parallel to the edge $D D 1$, their orthogonal projections onto the plane $\gamma$ are also equal to $\frac{\sqrt{2} \sqrt{3}}{\sqrt{3}}$. It is clear that the angle between the edge $C 1 D 1$ and the diagonal $B D 1$ is also $\alpha$. Therefore, the orthogonal projection of the edge $C 1 D 1$ onto the plane $\gamma$ is $C 1 D 1 \cos \left(90^{\circ}-\alpha\right)=a \sin \alpha=\frac{a \sqrt{2}}{\sqrt{3}}$. Since the edges $A B, A 1 B 1$ and $C D$ are equal and parallel to the edge $C 1 D 1$, their orthogonal projections onto the plane $\gamma$ are also equal to $\sqrt{\frac{a}{2}} \sqrt{3}$. Thus, the orthogonal projection of the cube onto the plane $\gamma$ is a hexagon, all sides of which are equal to $\frac{\sqrt{2} a}{\sqrt{3}}$. Since the diagonal $A C$ of the square $A B C D$ is perpendicular to the diagonal $B D 1$ of the cube, its orthogonal projection onto the plane $\gamma$ is parallel and equal to the segment $A C=a \sqrt{2}$. Let $A 2, B 2$ and $C 2$ be the orthogonal projections of the vertices $A, B$ and $C$ onto the plane $\gamma$ (Fig.3). Then in the triangle $A 2 B 2 C 2$, it is known that

$$
A 2 C 2=a \sqrt{2}, A 2 B 2=B 2 C 2=\frac{a \sqrt{2}}{\sqrt{3}}
$$

Therefore, $\angle A 2 B 2 C 2=120^{\circ}$. Similarly, all angles of the hexagon, which is the orthogonal projection of the cube $A B C D A 1 B 1 C 1 D 1$ onto the plane $\gamma$, are $120^{\circ}$. Consequently, this hexagon is regular.

## Answer

A regular hexagon with side $\frac{a \sqrt{2}}{\sqrt{3}}$, where $a$ is the edge of the cube.