Example 6 As shown in Figure 7, given a cube $A B C D$ $A_{1} B_{1} C_{1} D_{1}$. Any plane $\alpha$ is perpendicular to the diagonal $A C_{1}$, such that plane $\alpha$ intersects each face of the cube. Let the area of the resulting cross-sectional polygon be $S$, and the perimeter be $l$. Then ().
(A) $S$ is a constant, $l$ is not a constant
(B) Both $S$ and $l$ are constants
(C) $S$ is not a constant, $l$ is a constant
(D) Neither $S$ nor $l$ are constants
(2005, National High School Mathematics Competition)
Consider the special case first.
Assume the edge length of the cube is 1. As shown in Figure 7, take points $E$, $F$, $G$, $H$, $I$, and $J$ as the midpoints of the six edges. Clearly, the regular hexagon $E F G H I J$ is a valid cross-section, with its perimeter $l_{1}=3 \sqrt{2}$ and area $S_{1}=\frac{3 \sqrt{3}}{4}$. When the cross-section is the equilateral triangle $A_{1} B D$, its perimeter $l_{2}=3 \sqrt{2}$, and area $S_{2}=\frac{\sqrt{3}}{2}$.

Notice that $l_{1}=l_{2}$, but $S_{1} \neq S_{2}$. This indicates that $S$ is not a constant, while $l$ might be a constant.
Now consider the general case.
Let the hexagon $J_{1} E_{1} F_{1} G_{1} H_{1} I_{1}$ be any valid cross-section. This cross-section is parallel to the two special cross-sections mentioned earlier.
By the property of similar triangles, we have
$$
\begin{array}{l}
\frac{J_{1} E_{1}}{D_{1} B_{1}}=\frac{A_{1} E_{1}}{A_{1} B_{1}}, \frac{E_{1} F_{1}}{A_{1} B}=\frac{B_{1} E_{1}}{A_{1} B_{1}} . \\
\text { Therefore, } J_{1} E_{1}=\sqrt{2} A_{1} E_{1}, E_{1} F_{1}=\sqrt{2} B_{1} E_{1}, \\
J_{1} E_{1}+E_{1} F_{1}=\sqrt{2}\left(A_{1} E_{1}+B_{1} E_{1}\right) \\
=\sqrt{2} A_{1} B_{1}=\sqrt{2} .
\end{array}
$$

Similarly, the sum of the other four sides is $2 \sqrt{2}$.
Thus, the perimeter of the hexagon $J_{1} E_{1} F_{1} G_{1} H_{1} I_{1}$ is a constant $3 \sqrt{2}$.
Therefore, the correct choice is (C).