Three. (20 points) As shown in Figure 4, given that the edge length of the cube $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$ is $1, \odot O_{1}$ is the incircle of the square $A B C D$, $\odot O_{2}$ is the circumcircle of the square $A D D_{1} A_{1}$, and $P, Q$ are points on $\odot O_{1}$ and $\odot O_{2}$, respectively. Find the range of the length of $P Q$.
Three, establish the spatial rectangular coordinate system as shown in Figure 8.
In the $x 0 y$ plane,

the equation of $\odot O_{1}$ is
$$
\begin{array}{c}
\left(x-\frac{1}{2}\right)^{2}+ \\
\left(y-\frac{1}{2}\right)^{2}=\frac{1}{4}
\end{array}
$$

In the $x O_{z}$ plane, the equation of $\odot \mathrm{O}_{2}$ is
$$
\left(x-\frac{1}{2}\right)^{2}+\left(z-\frac{1}{2}\right)^{2}=\frac{1}{2} \text {. }
$$

Let $P\left(\frac{1}{2}+\frac{1}{2} \cos \theta, \frac{1}{2}+\frac{1}{2} \sin \theta, 0\right)$,
$$
Q\left(\frac{1}{2}+\frac{\sqrt{2}}{2} \cos \varphi, 0, \frac{1}{2}+\frac{\sqrt{2}}{2} \sin \varphi\right) \text {. }
$$

Then $|P Q|^{2}$
$$
\begin{array}{l}
=\left(\frac{1}{2} \cos \theta-\frac{\sqrt{2}}{2} \cos \varphi\right)^{2}+\left(\frac{1}{2}+\frac{1}{2} \sin \theta\right)^{2}+\left(\frac{1}{2}+\frac{\sqrt{2}}{2} \sin \theta\right)^{2} \\
=\frac{5}{4}+\frac{1}{2}(\sin \theta+\sqrt{2} \sin \varphi-\sqrt{2} \cos \theta \cdot \cos \varphi) \text {, } \\
M=\sin \theta+\sqrt{2} \cdot \sin \varphi-\sqrt{2} \cos \theta \cdot \cos \varphi \\
=\sin \theta+\sqrt{2} \cdot \sqrt{1+\cos ^{2} \theta} \text {. } \\
\left(\frac{1}{\sqrt{1+\cos ^{2} \theta}} \sin \varphi-\frac{\cos \theta}{\sqrt{1+\cos ^{2} \theta}} \cos \varphi\right) \\
=\sin \theta+\sqrt{2} \cdot \sqrt{1+\cos ^{2} \theta} \cdot \sin (\varphi-\alpha) \\
\leqslant \sin \theta+\sqrt{2} \cdot \sqrt{1+\cos ^{2} \theta} \\
\leqslant|\sin \theta|+\sqrt{2} \cdot \sqrt{1+\cos ^{2} \theta} \\
=|\sin \theta|+\frac{\sqrt{2}}{2} \cdot \sqrt{1+\cos ^{2} \theta}+\frac{\sqrt{2}}{2} \cdot \sqrt{1+\cos ^{2} \theta} \\
\leqslant 3 \sqrt{\frac{\sin ^{2} \theta+\left(\frac{\sqrt{2}}{2} \cdot \sqrt{1+\cos ^{2} \theta}\right)^{2} \cdot 2}{3}} \\
=\sqrt{6} \text {. } \\
\end{array}
$$

where, $\alpha=\arccos \frac{1}{\sqrt{1+\cos ^{2} \theta}}=\arcsin \frac{\cos \theta}{\sqrt{1+\cos ^{2} \theta}}$.
When $\theta \stackrel{\ell}{=} \arcsin \frac{\sqrt{6}}{3}, \varphi=\frac{2 \pi}{3}$, the equality in (1) holds.
On the other hand,
$$
\begin{array}{l}
M \geqslant \sin \theta-\sqrt{2} \cdot \sqrt{1+\cos ^{2} \theta} \\
\geqslant-|\sin \theta|-\sqrt{2} \cdot \sqrt{1+\cos ^{2} \theta} \\
=-\left(|\sin \theta|+\sqrt{2} \cdot \sqrt{1+\cos ^{2} \theta}\right) \geqslant-\sqrt{6} .
\end{array}
$$

When $\theta=\pi+\arcsin \frac{\sqrt{6}}{3}, \varphi=\frac{5 \pi}{3}$, the equality in the above expression holds.
Therefore, $\frac{5}{4}-\frac{\sqrt{6}}{2} \leqslant|P Q|^{2} \leqslant \frac{5}{4}+\frac{\sqrt{6}}{2}$, i.e.,
$$
\frac{\sqrt{3}-\sqrt{2}}{2} \leqslant|P Q| \leqslant \frac{\sqrt{3}+\sqrt{2}}{2} \text {. }
$$