In Δ ABC,

Let $\frac{b+c}{12}=\frac{c+a}{13}=\frac{a+b}{15}=k$

$\Rightarrow b+c=12 k, c+a=13 k, a+b=15 k$

$\Rightarrow b+c+c+a+a+b=12 k+13 k+15 k$

$\Rightarrow 2(a+b+c)=40 k$

$\Rightarrow a+c+b=20 k$

$\Rightarrow a+12 k=20 k \quad(\because b+c=12 k)$

$\Rightarrow a=8 k$

Also,

$c+a=13 k$

$\Rightarrow c=13 k-a=13 k-8 k=5 k$

and,

$a+b=15 k$

$\Rightarrow b=15 k-a=15 k-8 k=7 k$

Now,

$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{k^{2}(49+25-64)}{k^{2}(2 \times 35)}=\frac{1}{7}$

$\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{k^{2}(64+25-49)}{(2 \times 40) k^{2}}=\frac{1}{2}$

$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{k^{2}(64+49-25)}{(2 \times 56) k^{2}}=\frac{11}{14}$

$\therefore \quad \cos A: \cos B: \cos C=\frac{1}{7}: \frac{1}{2}: \frac{11}{14}=2: 7: 11$

$\Rightarrow \frac{\cos A}{2}=\frac{\cos B}{7}=\frac{\cos C}{11}$

Hence proved.