According to the condition, we can set
$$P_1(1, \cos \alpha_1, \sin \alpha_1), \quad P_2(\sin \alpha_2, 1, \cos \alpha_2), \quad P_3(\cos \alpha_3, \sin \alpha_3, 1).$$
We conventionally assume that $P_4 = P_1$ and $\alpha_4 = \alpha_1$. Denote
$$d_i = |P_i P_{i+1}| \quad (i = 1, 2, 3).$$
Then
$$d_i^2 = (1 - \sin \alpha_{i+1})^2 + (1 - \cos \alpha_i)^2 + (\sin \alpha_i - \cos \alpha_{i+1})^2.$$
Denote $f = |P_1P_2| + |P_2P_3| + |P_3P_1|$.

We first find the minimum of $f$. For $i = 1, 2, 3$, according to (1) and the inequality of arithmetic and geometric means we get
$$d_i^2 \geq (1 - \sin \alpha_{i+1})^2 + (1 - \cos \alpha_i)^2 \\ \geq \frac{1}{2}((1 - \sin \alpha_{i+1}) + (1 - \cos \alpha_i))^2,$$
and thus $d_i \geq \frac{\sqrt{2}}{2}(2 - \sin \alpha_{i+1} - \cos \alpha_i)$. Therefore,
$$f = d_1 + d_2 + d_3 \\ \geq \frac{\sqrt{2}}{2} \sum_{i=1}^{3} (2 - \sin \alpha_{i+1} - \cos \alpha_i) \\ = 3\sqrt{2} - \frac{\sqrt{2}}{2} \sum_{i=1}^{3} (\sin \alpha_i + \cos \alpha_i) \\ = 3\sqrt{2} - \sum_{i=1}^{3} \sin \left(\alpha_i + \frac{\pi}{4}\right) \\ \geq 3\sqrt{2} - 3.$$
When $\alpha_i = \frac{\pi}{4}$ ($i = 1, 2, 3$), $f$ can take the minimum $3\sqrt{2} - 3$.

Then we will find the maximum of $f$. By (1), it is clear that
$$d_i^2 = 4 - 2 \cos \alpha_i - 2 \sin \alpha_{i+1} - 2 \sin \alpha_i \cos \alpha_{i+1}.$$
Note that $\sin \alpha_i \geq -1$, $\cos \alpha_i \geq -1$ ($i = 1, 2, 3$), and then
$$\begin{aligned} \sum_{i=1}^{3} d_i^2 &= 12 - 2 \left( \sum_{i=1}^{3} \sin \alpha_{i+1} + \sum_{i=1}^{3} \cos \alpha_i + \sum_{i=1}^{3} \sin \alpha_i \cos \alpha_{i+1} \right) \\ &= 12 - 2 \left( \sum_{i=1}^{3} \sin \alpha_i + \sum_{i=1}^{3} \cos \alpha_{i+1} + \sum_{i=1}^{3} \sin \alpha_i \cos \alpha_{i+1} \right) \\ &= 18 - 2 \sum_{i=1}^{3} (1 + \sin \alpha_i)(1 + \cos \alpha_{i+1}) \leq 18. \end{aligned}$$
By the Cauchy inequality, we know that $f^2 \leq 3(d_1^2 + d_2^2 + d_3^2) = 54$, so $f \leq 3\sqrt{6}$.

When $\alpha_i = \pi$ ($i = 1, 2, 3$), $f$ can take the maximum $3\sqrt{6}$. In conclusion, the minimum of $f$ is $3\sqrt{2} - 3$, and its maximum is $3\sqrt{6}$. $\square$