With usual notation, in a triangle ABC $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then the value of…

Let the common ratio be $k$, giving the system:

$\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13} = k$

$b + c = 11k$
$c + a = 12k$
$a + b = 13k$

Adding yields:

$(b + c) + (c + a) + (a + b) = 11k + 12k + 13k$
$2(a + b + c) = 36k$
$a + b + c = 18k$

Solving for each variable:

$a = (a + b + c) - (b + c) = 18k - 11k = 7k$
$b = (a + b + c) - (a + c) = 18k - 12k = 6k$
$c = (a + b + c) - (a + b) = 18k - 13k = 5k$

Applying the Law of Cosines:

$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$

$\cos B = \frac{(7k)^2 + (5k)^2 - (6k)^2}{2(7k)(5k)} = \frac{49k^2 + 25k^2 - 36k^2}{70k^2} = \frac{38k^2}{70k^2} = \frac{19}{35}$

The result corresponds to option C.

Final answer: $\boxed{\frac{19}{35}}$