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(C) Let $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k$. Then $b+c = 11k$,$c+a = 12k$,and $a+b = 13k$. Adding these equations,we get $2(a+b+c)

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$\frac{19}{35}$

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(C) Let $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k$. Then $b+c = 11k$,$c+a = 12k$,and $a+b = 13k$. Adding these equations,we get $2(a+b+c) = 36k$,so $a+b+c = 18k$. Subtracting the individual equations from the sum: $a = (a+b+c) - (b+c) = 18k - 11k = 7k$. $b = (a+b+c) - (c+a) = 18k - 12k = 6k$. $c = (a+b+c) - (a+b) = 18k - 13k = 5k$. Using the cosine rule,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$. Substituting the values: $\cos B = \frac{(7k)^2 + (5k)^2 - (6k)^2}{2(7k)(5k)} = \frac{49k^2 + 25k^2 - 36k^2}{70k^2} = \frac{38k^2}{70k^2} = \frac{19}{35}$.

With usual notation,in a triangle $ABC$,if $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$,then the value of $\cos B$ is equal to

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