Three, the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ is 1. On the face diagonal $A_{1} D$, take point $M$, and on $C D_{1}$, take point $N$, such that the line segment $M N$ is parallel to the diagonal plane $A_{1} A C C_{1}$. Calculate the minimum length of the line segment $M N$ that satisfies the above conditions.
Three, draw perpendiculars from $M$ and $N$ to the base, with the feet of the perpendiculars $M_{1}$ on $A D$ and $N_{1}$ on $C D$. Connect $M_{1} N_{1}$, then $M_{1} N_{1} \| A C$. As shown in the figure, let $M M_{1}=\boldsymbol{x}$, $N N_{1}=1-x$. Draw $M H \perp N N_{1}$ at $H$, then $N H=1-2 x, M H=\sqrt{2} x$. Therefore, $M N^{2}$ $=6\left(x-\frac{1}{3}\right)^{2}+\frac{1}{3}$. When $x=\frac{1}{3}$, the minimum value of $M N^{2}$ is $\frac{1}{3}$. Hence, the minimum value of $M N$ is $\frac{\sqrt{3}}{3}$.