Law of Cosines

$\Large{\sum_{cyc} \cos 3A = 1 - 4\prod_{cyc} \sin \dfrac{3A}{2} \\ 1 = 1 - 4\prod_{cyc} \sin \dfrac{3A}{2} \\ \Rightarrow \prod_{cyc} \sin \dfrac{3A}{2} = 0}$

This means one of the sines of $\dfrac{3A}{2},\dfrac{3B}{2},\dfrac{3C}{2}$ is $0$ .

Suppose $\sin \dfrac{3A}{2}=0 \Rightarrow \dfrac{3A}{2} = 180 \ \because \sin(180)=0 \Rightarrow 3A = 360 \Rightarrow A=120$

Thus , the longest side is the one opposite the obtuse angle i.e $120$ .

Using cosine rule ,

$\Large{(\sqrt{m})^2=10^2 + 13^2 -2(10)(13)\cos 120 \\ \Rightarrow m=10^2 + 13^2 -2(10)(13)\left(\dfrac{-1}{2}\right) \\ \Rightarrow m=100+169+130 = \boxed{399}}$

As above, we can see that $\cos3A+\cos3B-\cos(3A+3B)=1$

Expanding, we get

$\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1$

$\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B$

$(\cos3A-1)(\cos3B-1)=\sin3A\sin3B$

$\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1$

$\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1$

Note that $\tan{x}=\frac{1}{\tan(90-x)}$ , or $\tan{x}\tan(90-x)=1$

Thus $\frac{3A}{2}+\frac{3B}{2}=90$ , or $A+B=60$ .

Now we know that $C=120$ , so we can just use Law or Cosines to get $\boxed{399}$