10. (20 points) In $\triangle A B C$, the lengths of the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively,
$$
\begin{array}{l}
\angle C=\max \{\angle A, \angle B, \angle C\}, \\
\sin C=1+\cos C \cdot \cos (A-B), \\
\frac{2}{a}+\frac{1}{b}=1 .
\end{array}
$$

Find the minimum perimeter of $\triangle A B C$.
10. Note that,
$$
\begin{array}{l}
\sin C=1+\cos C \cdot \cos (A-B) \\
=1-\cos (A+B) \cdot \cos (A-B) \\
=1-\frac{1}{2}(\cos 2 A+\cos 2 B) \\
=\sin ^{2} A+\sin ^{2} B \\
\Rightarrow \sin ^{2} A+\sin ^{2} B=\sin C \\
=\sin A \cdot \cos B+\sin B \cdot \cos A .
\end{array}
$$

When $\angle A+\angle B\sin B, \cos B>\sin A$
$\Rightarrow \sin ^{2} A+\sin ^{2} B\frac{\pi}{2}$, we have
$$
\begin{array}{l}
\cos A\sin A \cdot \cos B+\sin B \cdot \cos A .
\end{array}
$$

When $\angle A+\angle B=\frac{\pi}{2}$, we have
$\triangle A B C$'s perimeter $=a+b+c$
$$
\begin{array}{l}
=a+b+\sqrt{a^{2}+b^{2}} \\
=a+b+\sqrt{\left(a^{2}+b^{2}\right)\left(\left(\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}\right)} \\
\geqslant a+b+\frac{4 a}{5}+\frac{3 b}{5}=\frac{1}{5}(9 a+8 b)\left(\frac{2}{a}+\frac{1}{b}\right) \\
\geqslant \frac{1}{5}(3 \sqrt{2}+2 \sqrt{2})^{2}=10 .
\end{array}
$$

Therefore, when $a=\frac{10}{3}, b=\frac{5}{2}$, the minimum perimeter of $\triangle A B C$ is 10.