11. As shown in the figure, the edge length of the cube $A B C D-E F G H$ is 2. Take any point $P_{1}$ on the incircle of the square $A B F E$, any point $P_{2}$ on the incircle of the square $B C G F$, and any point $P_{3}$ on the incircle of the square $E F G H$. Find the minimum and maximum values of $\left|P_{1} P_{2}\right|+\left|P_{2} P_{3}\right|+\left|P_{3} P_{1}\right|$.
Solve:
As shown in the figure, with $\overrightarrow{D A}, \overrightarrow{D B}, \overrightarrow{D H}$ as the positive directions of the $x, y, z$ axes, establish a Cartesian coordinate system.
$$
\begin{array}{l}
P_{1}\left(2,1+\cos \alpha_{1}, 1+\sin \alpha_{1}\right), \\
P_{2}\left(1+\cos \alpha_{2}, 2,1+\sin \alpha_{2}\right), \\
P_{3}\left(1+\cos \alpha_{3}, 1+\sin \alpha_{3}, 2\right),
\end{array}
$$

Then we have $\left|P_{1} P_{2}\right|^{2}=\left(1-\cos \alpha_{2}\right)^{2}+\left(1-\cos \alpha_{1}\right)^{2}$
$$
\begin{array}{l}
+\left(\sin \alpha_{1}-\sin \alpha_{2}\right)^{2}, \\
\left|P_{1} P_{3}\right|^{2}=\left(1-\cos \alpha_{3}\right)^{2}+\left(\cos \alpha_{1}-\sin \alpha_{3}\right)^{2}+\left(1-\sin \alpha_{1}\right)^{2}, \\
\left|P_{2} P_{3}\right|^{2}=\left(\cos \alpha_{2}-\cos \alpha_{3}\right)^{2}+\left(1-\sin \alpha_{3}\right)^{2}+\left(1-\sin \alpha_{2}\right)^{2} .
\end{array}
$$

Thus, $\left|P_{1} P_{2}\right|^{2}+\left|P_{1} P_{3}\right|^{2}+\left|P_{2} P_{3}\right|^{2}=4-2\left(\cos \alpha_{2}+\cos \alpha_{1}+\sin \alpha_{1} \sin \alpha_{2}\right)$
$$
\begin{array}{l}
+4-2\left(\cos \alpha_{3}+\sin \alpha_{1}+\cos \alpha_{1} \sin \alpha_{3}\right)+4-2\left(\sin \alpha_{3}+\sin \alpha_{2}+\cos \alpha_{2} \cos \alpha_{3}\right) \\
=18-2\left[\left(1+\sin \alpha_{1}\right)\left(1+\sin \alpha_{2}\right)+\left(1+\cos \alpha_{1}\right)\left(1+\sin \alpha_{3}\right)\right.
\end{array}
$$
$\left.+\left(1+\cos \alpha_{2}\right)\left(1+\cos \alpha_{3}\right)\right] \leqslant 18$, when $\alpha_{1}=\alpha_{3}=-\frac{\pi}{2}, \alpha_{2}=\pi$ the equality holds.
Therefore, $\left(\left|P_{1} P_{2}\right|+\left|P_{2} P_{3}\right|+\left|P_{3} P_{1}\right|\right)^{2} \leqslant 3\left(\left|P_{1} P_{2}\right|^{2}+\left|P_{1} P_{3}\right|^{2}+\left|P_{2} P_{3}\right|^{2}\right) \leqslant 54$
$$
\Rightarrow\left|P_{1} P_{2}\right|+\left|P_{2} P_{3}\right|+\left|P_{3} P_{1}\right| \leqslant 3 \sqrt{6} \text {. }
$$

On the other hand, $\left|P_{1} P_{2}\right|^{2} \geqslant\left(1-\cos \alpha_{2}\right)^{2}+\left(1-\cos \alpha_{1}\right)^{2} \geqslant \frac{1}{2}\left(2-\cos \alpha_{2}-\cos \alpha_{1}\right)^{2}$ $\Rightarrow\left|P_{1} P_{2}\right| \geqslant \frac{\sqrt{2}}{2}\left(2-\cos \alpha_{2}-\cos \alpha_{1}\right)$, similarly we get
$$
\left|P_{1} P_{3}\right| \geqslant \frac{\sqrt{2}}{2}\left(2-\cos \alpha_{3}-\sin \alpha_{1}\right),\left|P_{2} P_{3}\right| \geqslant \frac{\sqrt{2}}{2}\left(2-\sin \alpha_{3}-\sin \alpha_{2}\right) \text {. }
$$

Thus, $\left|P_{1} P_{2}\right|+\left|P_{2} P_{3}\right|+\left|P_{3} P_{1}\right|$
$$
\geqslant 3 \sqrt{2}-\left[\sin \left(\alpha_{1}+\frac{\pi}{4}\right)+\sin \left(\alpha_{2}+\frac{\pi}{4}\right)+\sin \left(\alpha_{3}+\frac{\pi}{4}\right)\right] \geqslant 3 \sqrt{2}-3,
$$

when $\alpha_{1}=\alpha_{2}=\alpha_{3}=\frac{\pi}{4}$ the equality holds.
In summary, the minimum value of $\left|P_{1} P_{2}\right|+\left|P_{2} P_{3}\right|+\left|P_{3} P_{1}\right|$ is $3 \sqrt{2}-3$, and the maximum value is $3 \sqrt{6}$.