14. Let $a, b, c$ be positive real numbers such that
$$
\left\{\begin{array}{l}
a^{2}+a b+b^{2}=25, \\
b^{2}+b c+c^{2}=49, \\
c^{2}+a+a^{2}=64 .
\end{array}\right.
$$

Find $(a+b+c)^{2}$.
14. Answer: 129.
Consider the following picture, where $\angle A O B=\angle B O C=\angle C O A=120^{\circ}, O A=a, O B=b$ and $O C=c$.

Then $|B C|=5,|C A|=7$ and $|A B|=8$. The area of the triangle $A B C$ is
$$
\sqrt{10(10-5)(10-7)(10-8)}=10 \sqrt{3} \text {. }
$$

Then $\frac{1}{2} \frac{\sqrt{3}}{2}(a b+b c+c a)=10 \sqrt{3}$. So $a b+b c+c a=40$.
$$
2(a+b+c)^{2}=\left(a^{2}+a b+b^{2}\right)+\left(b^{2}+b c+c^{2}\right)+\left(c^{2}+c a+a^{2}\right)+3(a b+b c+c a)=258 .
$$

Thus, $(a+b+c)^{2}=129$.