Three, as shown in Figure 1, let $\square E B E_{1} D_{1}$ be the section through the diagonal $B D_{1}$ of the cube. Then $S_{\square E E E_{1} D}=S=2 S_{\triangle A D_{1} E_{1}}=h B D_{1}$, where $h$ is the distance from $E_{1}$ to $B D_{1}$. When $S$ is minimized, $h$ also takes its minimum value $h^{\prime}$. It is easy to see that $h^{\prime}$ is the distance between the skew lines $B D_{1}$ and $B_{1} C_{1}$, which is also equal to the distance from $B_{1}$ to the plane $A_{1} B D_{1}$. Therefore,
$$\begin{array}{l} V_{\text {truncated } B_{1} \cdot A_{1} H D_{1}}=\frac{1}{3} h^{\prime}\left(\frac{1}{2} \sqrt{2} a^{2}\right)=\frac{\sqrt{2}}{6} h^{\prime} a^{2}, \\ V_{\text {truncated } D_{1} \cdot B_{1} A_{1} B}=\frac{1}{3} a\left(\frac{1}{2} a^{2}\right)=\frac{a^{3}}{6}. \end{array}$$

Here $a$ is the edge length of the cube. From this, we get
$$\frac{\sqrt{2}}{6} h^{\prime} a^{2}=\frac{a^{3}}{6}.$$

Thus, $h^{\prime}=\frac{a}{\sqrt{2}}$, and
$$S_{\text {minimum }}=h^{\prime} \cdot B D_{1}=\frac{a}{\sqrt{2}} \cdot \sqrt{3} a=\frac{\sqrt{6}}{2} a^{2}.$$

It is easy to see that $S_{\text {maximum }}=S_{\square D B B_{1} D_{1}}=\sqrt{2} c^{2}$.