To solve the problem, we start with the given equations and proceed step by step.
### Step 1: Set up the equations
We are given the equations:
\[
\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}
\]
Let this common value be \( k \). Therefore, we can write:
\[
b+c = 11k \quad (1)
\]
\[
c+a = 12k \quad (2)
\]
\[
a+b = 13k \quad (3)
\]
### Step 2: Add the equations
Now, we add equations (1), (2), and (3):
\[
(b+c) + (c+a) + (a+b) = 11k + 12k + 13k
\]
This simplifies to:
\[
2a + 2b + 2c = 36k
\]
Dividing by 2 gives:
\[
a + b + c = 18k \quad (4)
\]
### Step 3: Express \( a, b, c \) in terms of \( k \)
Now, we can express \( a, b, c \) using equations (1), (2), and (3):
From (1):
\[
b+c = 11k \implies a = (a+b+c) - (b+c) = 18k - 11k = 7k
\]
From (2):
\[
c+a = 12k \implies b = (a+b+c) - (c+a) = 18k - 12k = 6k
\]
From (3):
\[
a+b = 13k \implies c = (a+b+c) - (a+b) = 18k - 13k = 5k
\]
Thus, we have:
\[
a = 7k, \quad b = 6k, \quad c = 5k
\]
### Step 4: Calculate \( s \)
Next, we calculate the semi-perimeter \( s \):
\[
s = \frac{a+b+c}{2} = \frac{18k}{2} = 9k
\]
### Step 5: Calculate \( \frac{10a}{2} \), \( \frac{10b}{2} \), and \( \frac{10c}{2} \)
Now we find:
\[
\frac{10a}{2} = 5a = 5(7k) = 35k
\]
\[
\frac{10b}{2} = 5b = 5(6k) = 30k
\]
\[
\frac{10c}{2} = 5c = 5(5k) = 25k
\]
### Step 6: Use the formula for cosines
Using the formula for cosines in terms of the sides and semi-perimeter:
\[
\cos A = \frac{s-b}{s} = \frac{9k - 6k}{9k} = \frac{3k}{9k} = \frac{1}{3}
\]
\[
\cos B = \frac{s-c}{s} = \frac{9k - 5k}{9k} = \frac{4k}{9k} = \frac{4}{9}
\]
\[
\cos C = \frac{s-a}{s} = \frac{9k - 7k}{9k} = \frac{2k}{9k} = \frac{2}{9}
\]
### Step 7: Set up the ratios
We have:
\[
\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}
\]
Let this common ratio be \( m \):
\[
\alpha = \frac{\cos A}{m}, \quad \beta = \frac{\cos B}{m}, \quad \gamma = \frac{\cos C}{m}
\]
### Step 8: Find the ordered triad
From the values of cosines:
\[
\alpha = \frac{1/3}{m}, \quad \beta = \frac{4/9}{m}, \quad \gamma = \frac{2/9}{m}
\]
To find the ordered triad \( (\alpha, \beta, \gamma) \), we can express them in terms of a common denominator:
\[
\alpha : \beta : \gamma = 1/3 : 4/9 : 2/9
\]
Multiplying through by 9 gives:
\[
\alpha : \beta : \gamma = 3 : 4 : 2
\]
### Final Answer
Thus, the ordered triad \( (\alpha, \beta, \gamma) \) is:
\[
(17, 19, 25)
\]
