3. For the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, the area of the section through the diagonal $B D_{1}$ is $S$, then the value of $S_{\text {max }} / S_{\text {min }}$ is ( ).
A. $\frac{\sqrt{3}}{2}$
B. $\frac{\sqrt{6}}{2}$
C. $\frac{2 \sqrt{3}}{3}$
D. $\frac{2 \sqrt{6}}{3}$
3. C

As shown in the figure, let the section through $B D_{1}$ be $\square B E D_{1} E_{1}$, then $S^{\prime}=2 S_{\triangle B D_{1} E_{1}}=h \cdot B D_{1}$, where $h$ is the distance from $E_{1}$ to $B D_{1}$. It is easy to see that $h$ is the minimum when it is the distance $h^{\prime}$ between the skew lines $B D_{1}$ and $B_{1} C_{1}$, at this time $h=h^{\prime}=\frac{a}{\sqrt{2}}$ (where $a$ is the edge length of the cube), thus $S_{\min }=\frac{a}{\sqrt{2}} \cdot \sqrt{3} a-\frac{\sqrt{6}}{2} a^{2}$. Also, $S_{\operatorname{max}}-S_{\triangle H B_{1} D_{1} D}=\sqrt{2} a^{2}$, so $S_{\max } / S_{\operatorname{max}}=\frac{2 \sqrt{3}}{3}$.