The angles of triangle ABC satisfy $\cos 3A+\cos 3B+\cos 3C=1$. Two sides of the triangle have lengths 10 and 13. The maximum possible length of the remaining side of  $\Delta \mathcal{l}e$ABC is ______

$$\begin{gathered} \cos 3A+\cos 3B=1-\cos 3C=1-\cos (3\pi -(3A+3B)) \\ \cos 3A+\cos 3B=1+\cos (3A+3B) \\ 2\cos \left(\frac{3A+3B}{2}\right)\cos \left(\frac{3A-3B}{2}\right)=2{\cos }^2\left(\frac{3(A+B)}{2}\right) \\ \Rightarrow \cos \frac{3(A+B)}{2}=0 or \Rightarrow 2\sin \frac{3A}{2}\sin \frac{3B}{2}=0 \\ \Rightarrow A+B=\frac{\pi }{3}\Rightarrow \overline{)C}=\frac{2\pi }{3} or \Rightarrow \frac{3A}{2}=\pi or \frac{3B}{2}=\pi \end{gathered}$$

$A=\frac{2\pi }{3} or B=\frac{2\pi }{3}$

maximum length of third side is possible when the two given sides are at an angle of  $\frac{2\pi }{3}$

$$\begin{gathered} {\mathcal{l}}^2={10}^2+{13}^2-\left(2\right)\left(10\right)\left(13\right)\left(\frac{-1}{2}\right)=100+109+130=399 \\ \mathcal{l}=\sqrt{399} \end{gathered}$$