(7) Given that $S$ is a parallelepiped, proposition $p$: “$S$ is a cube”, proposition $q$: “any diagonal of $S$ is perpendicular to the face diagonal it does not intersect”, then proposition $p$ is a ( ) of proposition $q$.
(A) sufficient but not necessary condition
(B) necessary but not sufficient condition
(C) sufficient and necessary condition
(D) neither sufficient nor necessary condition
$7 \mathrm{C}$ Sufficiency: It is obviously true, proof omitted.
Necessity: As shown in Figure (1), the top and bottom faces are parallelograms, and each side face is a rectangle.

Draw the projections of diagonals $A_{1} C$ and $B_{1} D$ on the plane $A A_{1} B_{1} B$. Since the plane $A A_{1} B_{1} B \perp$ plane $A B C D$, the feet of the perpendiculars from points $C$ and $D$ to the plane, $C^{\prime}$ and $D^{\prime}$, must lie on the line $A B$. The projections of $A_{1} C$ and $B_{1} D$ on the plane $A A_{1} B_{1} B$ are $A_{1} C^{\prime}$ and $B_{1} D^{\prime}$, respectively.
(Problem 7)
The figure on the plane $A A_{1} B_{1} B$ is shown in Figure (2). Since $C^{\prime} D^{\prime} \parallel A_{1} B_{1}$ and $C^{\prime} D^{\prime} = C D = C_{1} D_{1} = A_{1} B_{1}$, $A_{1} B_{1} C^{\prime} D^{\prime}$ is a parallelogram, and the intersection of its diagonals is point $Q$. By the theorem of three perpendiculars, $A_{1} C^{\prime} \perp A B_{1}$ and $B_{1} D^{\prime} \perp A_{1} B$. Therefore, $\angle A_{1} Q B_{1} + \angle A_{1} P B_{1} = \pi$.

The distance from point $Q$ to $A_{1} B_{1}$ is equal to the distance from point $P$ to $A_{1} B_{1}$, which is $\frac{1}{2} A A_{1}$. Here, point $P$ is the intersection of $A B_{1}$ and $A_{1} B$, so $S_{\triangle A_{1} P B_{1}} = S_{\triangle A_{1} Q B_{1}}$.

Since point $Q$ is the intersection of the altitudes from the two equal sides of the isosceles triangle $A_{1} P B_{1}$, $\triangle A_{1} Q B_{1}$ is also an isosceles triangle, thus
$$
\left|A_{1} P\right|^{2} \sin \angle A_{1} P B_{1} = \left|A_{1} Q\right|^{2} \sin \angle A_{1} Q B_{1}.
$$

This implies $A_{1} P = A_{1} Q$. Therefore, point $Q$ coincides with point $P$. And $\angle A_{1} P B_{1} = \angle A_{1} Q B_{1} = \frac{\pi}{2}$, so the side face $A A_{1} B_{1} B$ is a square. Similarly, it can be shown that the other side faces are also squares. Since points $D^{\prime}$ and $C^{\prime}$ coincide with points $A$ and $B$, respectively, the bottom face is a square.
In conclusion, $S$ is a cube.