II. (40 points) Let positive real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a+b=\sqrt{a b+9}, \\
b+c=\sqrt{b c+16}, \\
c+a=\sqrt{c a+25} .
\end{array}
$$

Find $a+b+c$.
$$
\begin{array}{l}
a^{2}+b^{2}-2 a b \cos 120^{\circ}=9, \\
b^{2}+c^{2}-2 b c \cos 120^{\circ}=16, \\
c^{2}+a^{2}-2 c a \cos 120^{\circ}=25 .
\end{array}
$$

From the cosine rule, we can construct the following geometric model.
In the plane, line segments $PA$, $PB$, and $PC$ share a common endpoint $P$ and the angles between each pair of segments are $120^{\circ}$, with $PA=a$, $PB=b$, and $PC=c$.
Thus, $AB^{2}=9$, $BC^{2}=16$, and $CA^{2}=25$.
Therefore, $\triangle ABC$ is a right triangle, and its area is 6.
$$
\begin{array}{l}
\text { Then } \frac{1}{2} a b \sin 120^{\circ}+\frac{1}{2} b c \sin 120^{\circ}+\frac{1}{2} c a \sin 120^{\circ}=6 \\
\Rightarrow a b+b c+c a=8 \sqrt{3} . \\
\text { Hence }(a+b+c)^{2} \\
=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a \\
=\frac{a^{2}+b^{2}+a b}{2}+\frac{b^{2}+c^{2}+b c}{2}+ \\
\frac{c^{2}+a^{2}+c a}{2}+\frac{3(a b+b c+c a)}{2} \\
=\frac{9}{2}+\frac{16}{2}+\frac{25}{2}+\frac{24 \sqrt{3}}{2} \\
=25+12 \sqrt{3} .
\end{array}
$$

Therefore, $a+b+c=\sqrt{25+12 \sqrt{3}}$.