Answer: 399

Solution:

Using $1 - \cos 3A = 2\sin^2\frac{3A}{2}$ and $\cos 3B + \cos 3C =$ $2\cos\frac{3(B+C)}{2}\cos\frac{3(B-C)}{2},$ together with $\frac{3(B+C)}{2} = 270^\circ - \frac{3A}{2}$ so that $\cos\frac{3(B+C)}{2} = -\sin\frac{3A}{2},$ the condition becomes $$\begin{aligned} 0 &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\sin\tfrac{3A}{2} + \cos\tfrac{3(B-C)}{2}\right) \\ &= 2\sin\tfrac{3A}{2} \\ &\quad {}\cdot \left(\cos\tfrac{3(B-C)}{2} - \cos\tfrac{3(B+C)}{2}\right) \\ &= 4\sin\tfrac{3A}{2}\sin\tfrac{3B}{2}\sin\tfrac{3C}{2}. \end{aligned}$$

For an angle $X$ of a triangle, $\frac{3X}{2}$ lies strictly between $0^\circ$ and $270^\circ,$ so $\sin\frac{3X}{2} = 0$ exactly when $X = 120^\circ.$ Hence one angle of the triangle is $120^\circ.$

The remaining side is longest when the $120^\circ$ angle sits between the sides of lengths $10$ and $13$ (if $120^\circ$ were opposite one of them, the remaining side would be shorter than that side). By the law of cosines its length is $\sqrt{10^2 + 13^2 + 10 \cdot 13} = \sqrt{399},$ so $m = 399.$