Let's determine the cube, given the following 7 points:

the $A_{1}$ and $A_{2}$ points on the bottom face plane,

the $F_{1}$ and $F_{2}$ points on the top face plane (the 4 points mentioned so far are not in the same plane),

a point $E$ on the front face plane,

a point $H$ on the back face plane,

a point $J$ on the right face plane.

(The terms "bottom, front, right, ..." were used only for easier orientation; originally, there is no designated vertical direction or viewing angle.)

What is the maximum number of such cubes?
The first four points clearly define the plane $S_{a}$ of the cube's lower face. Indeed, $S_{a}$ is parallel to the plane $S_{f}$ of the upper face, and thus also to the line $F_{1} F_{2}$, so the line $a_{1}$ passing through $A_{1}$ and parallel to $F_{1} F_{2}$ is also in $S_{a}$, hence $S_{a}$ is the plane passing through $a_{1}$ and $A_{2}$. ( $a_{1}$ cannot pass through $A_{2}$, otherwise the lines $A_{1} A_{2}$ and $F_{1} F_{2}$ would be parallel, lying in the same plane, which is excluded by the condition.) - Now $S_{f}$ is the plane passing through $F_{1}$ and parallel to $S_{a}$, the distance between them gives the length $a(>0)$ of the sought cube's edge, and the direction perpendicular to $S_{a}$ indicates the vertical direction.

The front edge $e$ of the cube's base passes through the projection $E^{\prime}$ of $E$ on $S_{a}$, the rear edge $h$ passes through the projection $H^{\prime}$ of $H$, and the distance between these two lines is $H^{\prime} H^{\prime \prime}=a$, where $H^{\prime \prime}$ is the projection of $H^{\prime}$ onto $e$. (These projections and the following $J^{\prime}$ are uniquely determined by $S_{a}$.) Thus, $E^{\prime} H^{\prime} H^{\prime \prime}$ is a right triangle, and $H^{\prime \prime}$ is the intersection of the Thales circle $k$ with diameter $E^{\prime} H^{\prime}$ and the circle $k_{h}$ of radius $a$ centered at $H^{\prime}$. Then the line $H^{\prime \prime} E^{\prime}$ gives us $e$, $h$ is parallel to $e$ and passes through $H^{\prime}$, and the right edge $j$ is perpendicular to $e$ and passes through the projection $J_{1}$ of the given point $J$ on $S_{a}$.

With this, we have obtained two vertices of the cube's base as the intersection points $J_{e}$ and $J_{h}$ of $j$ with $e$ and $h$, respectively, and the direction from $J_{e}$ to $J_{h}$ is clearly the viewing direction. Therefore, we must position ourselves on $S_{a}$ so that our head is on the same side of $S_{a}$ as $F_{1}$, facing in the direction of $j$, and such that $J_{e}$ is closer to us than $J_{h}$; this uniquely determines the "left" direction. Measuring the edge length $a$ from $J_{e}$ and $J_{h}$ in this direction, we uniquely obtain the vertices on the left side of the base, and finally, the projections of the base's 4 vertices on $S_{f}$ uniquely give the vertices of the cube at the back.

According to all this, exactly as many cubes meet the requirements as there are suitable points for the role of $H^{\prime \prime}$. Since the number of common points of $k$ and $k_{h}$ is at most 2, the number of suitable cubes is also at most 2. With this, we have answered the question of the problem.

(We can add: In the case $E^{\prime} H^{\prime} \geqq a$, 2 or 1 cube meets the requirements, in the case $E^{\prime} H^{\prime}<a$, there is no solution.)

Hajnal Imre (Szeged, Ságvári E. Gymnasium, I. o. t.) Borbély Albert (Budapest, Berzsenyi D. Gymnasium, II. o. t.)