In \(\triangle \mathrm{ABC}\), with usual notations, if \(\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}\), then the value of \(\cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}\) is

Let \(\frac{\mathrm{b}+\mathrm{c}}{11}=\frac{\mathrm{c}+\mathrm{a}}{12}=\frac{\mathrm{a}+\mathrm{b}}{13}=\mathrm{k}\)
\(\therefore \quad b+c=11 k\) ...(i)
\(c+a=12 k\) ...(ii)
and \(a+b=13 k\) ...(iii)
From (i) \(+(\) ii) + (iii), \(2(a+b+c)=36 k\)
\(\therefore \quad a+b+c=18 k\) ...(iv)
Now, (iv) - (i) gives, \(a=7 k\)
(iv) - (ii) gives, \(b=6 \mathrm{k}\)
(iv) - (iii) gives, \(c=5 \mathrm{k}\)
Now,
\(\begin{aligned} \cos A & =\frac{b^2+c^2-a^2}{2 b c} \\ & =\frac{(6 k)^2+(5 k)^2-(7 k)^2}{2 \times(6 k) \times(5 k)} \\ & =\frac{36 k^2+25 k^2-49 k^2}{60 k^2} \\ & =\frac{12 k^2}{60 k^2} \\ & =\frac{1}{5}\end{aligned}\)
\(\begin{aligned} \cos \mathrm{B} & =\frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ca}} \\ & =\frac{(5 \mathrm{k})^2+(7 \mathrm{k})^2-(6 \mathrm{k})^2}{2 \times(5 \mathrm{k}) \times(7 \mathrm{k})} \\ & =\frac{25 \mathrm{k}^2+49 \mathrm{k}^2-36 \mathrm{k}^2}{70 \mathrm{k}^2} \\ & =\frac{38 \mathrm{k}^2}{70 \mathrm{k}^2} \\ & =\frac{19}{35}\end{aligned}\)
\(\begin{aligned} \cos C & =\frac{\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2}{2 \mathrm{ab}}=\frac{(7 \mathrm{k})^2+(6 \mathrm{k})^2-(5 \mathrm{k})^2}{2 \times(7 \mathrm{k}) \times(6 \mathrm{k})} \\ & =\frac{49 \mathrm{k}^2+36 \mathrm{k}^2-25 \mathrm{k}^2}{84 \mathrm{k}^2}=\frac{60 \mathrm{k}^2}{84 \mathrm{k}^2}=\frac{5}{7}\end{aligned}\)
\(\therefore \quad \cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}=\frac{1}{5}+\frac{19}{35}+\frac{5}{7}\)
\(\begin{aligned} & =\frac{7+19+25}{35} \\ & =\frac{51}{35}\end{aligned}\)