Three. (20 points) For the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $2$, $M$ is the midpoint of edge $A A_{1}$. A circle $\Gamma$ is constructed through vertices $A$, $B_{1}$, and $C$. Let $P$ be any point on circle $\Gamma$. Find the range of values for the segment $P M$.

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Translation:

Three. (20 points) For the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $2$, $M$ is the midpoint of edge $A A_{1}$. A circle $\Gamma$ is constructed through vertices $A$, $B_{1}$, and $C$. Let $P$ be any point on circle $\Gamma$. Find the range of values for the segment $P M$.
Three, since $\triangle A B_{1} C$ is an equilateral triangle, let the body diagonal $B D_{1}$ intersect the plane of $\triangle A B_{1} C$ at point $O_{1}$, then $O_{1}$ is the center of circle $\Gamma$. It can be calculated that $B O_{1}=\frac{1}{3} B D_{1}, D_{1} O_{1}=\frac{2}{3} B D_{1}$.
As shown in Figure 6, with
$D_{1}$ as the origin, $D_{1} A_{1}$,
$D_{1} C_{1}, D_{1} D$ as the
positive directions of the $x$ axis, $y$ axis, $z$
axis, respectively, establish a
spatial rectangular coordinate system.
Then
$$
\begin{array}{l}
0_{1}\left(\frac{4}{3}, \frac{4}{3}, \frac{4}{3}\right), \\
M(2,0,1) .
\end{array}
$$

Thus, $\mathrm{O}_{1}$
$$
\begin{array}{l}
=\sqrt{\left(2-\frac{4}{3}\right)^{2}+\left(0-\frac{4}{3}\right)^{2}+\left(1-\frac{4}{3}\right)^{2}}=\frac{\sqrt{21}}{3}, \\
P O_{1}=A O_{1}=\frac{\sqrt{2}}{\cos 30^{\circ}}=\frac{2 \sqrt{6}}{3} .
\end{array}
$$

Let $\angle M O_{1} P=\theta$. In $\triangle M O_{1} P$, by the cosine rule we get
$$
\begin{array}{l}
M P=\sqrt{M O_{1}^{2}+P O_{1}^{2}-2 M O_{1} \cdot P O_{1} \cos \theta} \\
=\frac{1}{3} \sqrt{45-12 \sqrt{14} \cos \theta} .
\end{array}
$$

Clearly, the minimum value of $\theta$ is the angle (acute angle) $\theta_{0}$ between $M O_{1}$ and the plane $A B_{1} C$, and the maximum value of $\theta$ is $180^{\circ}-\theta_{0}$.
Next, we find the value of $\cos \theta_{0}$.
Clearly, $D_{1} B \perp$ plane $A B_{1} C$.
$$
\begin{array}{l}
\text { Also, } O_{1} M=\left(\frac{2}{3},-\frac{4}{3},-\frac{1}{3}\right), \\
a=3 O_{1} M=(2,-4,-1), \\
b=\frac{3}{4} O_{1} D_{1}=(-1,-1,-1),
\end{array}
$$

Then $\cos \left(90^{\circ}-\theta_{0}\right)=\frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}| \cdot|\boldsymbol{b}|}=\frac{1}{\sqrt{7}}$, that is
$$
\begin{array}{l}
\sin \theta_{0}=\frac{1}{\sqrt{7}}, \cos \theta_{0}=\sqrt{\frac{6}{7}} . \\
\cos \left(180^{\circ}-\theta_{0}\right)=-\sqrt{\frac{6}{7}} .
\end{array}
$$

Therefore, $M P \geqslant \frac{1}{3} \sqrt{45-12 \sqrt{14} \times \sqrt{\frac{6}{7}}}$
$$
\begin{array}{l}
=\frac{1}{3} \sqrt{45-24 \sqrt{3}}, \\
M P \leqslant \frac{1}{3} \sqrt{45+24 \sqrt{3}} .
\end{array}
$$

Thus, the range of $M P$ is
$$
\left[\frac{\sqrt{45-24 \sqrt{3}}}{3}, \frac{\sqrt{45+24 \sqrt{3}}}{3}\right] \text {. }
$$