Given $$\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$$ for a $$\Delta ABC$$ with usual notation. If $$\frac{\cos A}{a} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}$$, then the ordered triad $$(\alpha, \beta, \gamma)$$ has a value:

We are told that in the usual notation of $$\Delta ABC$$ the numbers $$a,\,b,\,c$$ (the side-lengths opposite to the angles $$A,\,B,\,C$$ respectively) satisfy

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

Because all three fractions are equal, we introduce a common value $$k$$ and write

$$b+c = 11k,\qquad c+a = 12k,\qquad a+b = 13k. \quad -(1)$$

Adding the three equalities in (1) we obtain

$$\bigl[(b+c)+(c+a)+(a+b)\bigr]=11k+12k+13k,$$

$$2(a+b+c)=36k,$$

$$a+b+c = 18k. \quad -(2)$$

To isolate each side separately we make simple combinations of the relations in (1).

• Add the second and the third and subtract the first:

$$\bigl[(c+a)+(a+b)-(b+c)\bigr] = (12k+13k-11k),$$

$$2a = 14k \;\Longrightarrow\; a = 7k. \quad -(3)$$

• Add the first and the third and subtract the second:

$$\bigl[(b+c)+(a+b)-(c+a)\bigr] = (11k+13k-12k),$$

$$2b = 12k \;\Longrightarrow\; b = 6k. \quad -(4)$$

• Add the first and the second and subtract the third:

$$\bigl[(b+c)+(c+a)-(a+b)\bigr] = (11k+12k-13k),$$

$$2c = 10k \;\Longrightarrow\; c = 5k. \quad -(5)$$

Thus the three sides are in the ratio

$$a:b:c = 7k:6k:5k,$$

or simply $$a=7k,\; b=6k,\; c=5k.$$

Next we need the cosines of the angles. By the Cosine Rule, which states $$\cos A=\dfrac{b^{2}+c^{2}-a^{2}}{2bc},\quad \cos B=\dfrac{c^{2}+a^{2}-b^{2}}{2ca},\quad \cos C=\dfrac{a^{2}+b^{2}-c^{2}}{2ab},$$ we substitute the obtained side-lengths.

For $$\cos A$$:

$$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc} =\frac{(6k)^{2}+(5k)^{2}-(7k)^{2}}{2\cdot6k\cdot5k} =\frac{36k^{2}+25k^{2}-49k^{2}}{60k^{2}} =\frac{12k^{2}}{60k^{2}} =\frac15. \quad -(6)$$

For $$\cos B$$:

$$\cos B=\frac{c^{2}+a^{2}-b^{2}}{2ca} =\frac{(5k)^{2}+(7k)^{2}-(6k)^{2}}{2\cdot5k\cdot7k} =\frac{25k^{2}+49k^{2}-36k^{2}}{70k^{2}} =\frac{38k^{2}}{70k^{2}} =\frac{19}{35}. \quad -(7)$$

For $$\cos C$$:

$$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab} =\frac{(7k)^{2}+(6k)^{2}-(5k)^{2}}{2\cdot7k\cdot6k} =\frac{49k^{2}+36k^{2}-25k^{2}}{84k^{2}} =\frac{60k^{2}}{84k^{2}} =\frac57. \quad -(8)$$

Now the problem tells us that

$$\frac{\cos A}{a}=\frac{\cos B}{\beta}=\frac{\cos C}{\gamma}.$$

Let the common value of these three fractions be $$\lambda$$. Then

$$\lambda=\frac{\cos A}{a} =\frac{\tfrac15}{7k} =\frac1{35k}. \quad -(9)$$

Using the same $$\lambda$$ for the other two ratios we can obtain $$\beta$$ and $$\gamma$$.

From $$\lambda=\dfrac{\cos B}{\beta}$$ we have

$$\beta=\frac{\cos B}{\lambda} =\frac{\tfrac{19}{35}}{\tfrac1{35k}} =\frac{19}{35}\cdot35k =19k. \quad -(10)$$

Similarly, from $$\lambda=\dfrac{\cos C}{\gamma}$$ we get

$$\gamma=\frac{\cos C}{\lambda} =\frac{\tfrac57}{\tfrac1{35k}} =\frac57\cdot35k =25k. \quad -(11)$$

Finally, to write the ordered triad $$(\alpha,\beta,\gamma)$$ completely we notice that, in perfect analogy with (10) and (11),

$$\alpha=\frac{\cos A}{\lambda} =\frac{\tfrac15}{\tfrac1{35k}} =7k. \quad -(12)$$

Therefore, neglecting the common positive factor $$k,$$ the required triple is

$$(\alpha,\beta,\gamma)=(7,\,19,\,25).$$

Hence, the correct answer is Option A.