The edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is $a$, and it is cut by a plane through its diagonal $A C_{1}$. The minimum value of the area of the section is $\qquad$ .
7. $\frac{\sqrt{6} a^{2}}{2}$.

As shown in the figure, the section must be a parallelogram, let it be $A M C_{1} N$. Draw $M K \perp A C_{1}$ with the foot of the perpendicular at $K$. Then the area of the parallelogram $A M C_{1} N$ is $S = M K \cdot A C_{1}$.

Since $A C_{1} = \sqrt{3} a$ is a fixed length, $S$ is minimized when $M K$ is the shortest. At this time, $M K$ is the common perpendicular segment length of the skew lines $B B_{1}$ and $A C_{1}$, which means $M$ is the midpoint of $B B_{1}$ (and thus $N$ is the midpoint of $D D_{1}$), and $M K = \frac{\sqrt{2} a}{2}$.
Therefore, $S_{\text {min }} = \sqrt{3} a \left(\frac{\sqrt{2} a}{2}\right) = \frac{\sqrt{6} a^{2}}{2}$.