Problem 1231

1. Given the equation:
$$\frac{a+b}{a+c} = \frac{b+c}{b+a}$$
Cross-multiplying, we get:
$$(a+b)(b+a) = (b+c)(a+c)$$
Simplifying, we have:
$$(a+b)^2 = (b+c)(a+c)$$

2. Expanding both sides, we get:
$$a^2 + 2ab + b^2 = ab + ac + bc + c^2$$
Rearranging terms, we have:
$$a^2 + 2ab + b^2 - ab - ac - bc - c^2 = 0$$
Simplifying further:
$$a^2 + ab + b^2 - ac - bc - c^2 = 0$$

3. Given that $ab + ac + bc$ is a prime number, we can rewrite the equation as:
$$(a+b+c)(a+b-c) = ab + ac + bc$$
Since $ab + ac + bc$ is a prime number, one of the factors must be 1. Therefore, we have two cases:
$$a+b+c = 1 \quad \text{or} \quad a+b-c = 1$$

4. Since $a, b, c$ are positive integers, $a+b+c = 1$ is not possible. Thus, we have:
$$a+b-c = 1$$
Solving for $c$, we get:
$$c = a + b - 1$$

5. Substituting $c = a + b - 1$ into the equation $(a+b)^2 = (b+c)(a+c)$, we get:
$$(a+b)^2 = (b + (a+b-1))(a + (a+b-1))$$
Simplifying, we have:
$$(a+b)^2 = (2a + b - 1)(2b + a - 1)$$

6. Expanding both sides, we get:
$$a^2 + 2ab + b^2 = (2a + b - 1)(2b + a - 1)$$
Expanding the right-hand side:
$$a^2 + 2ab + b^2 = 4ab + 2a^2 + 2b^2 - 2a - 2b + 1$$
Simplifying, we have:
$$a^2 + 2ab + b^2 = 2a^2 + 4ab + 2b^2 - 2a - 2b + 1$$
Rearranging terms, we get:
$$0 = a^2 + 2ab + b^2 - 2a^2 - 4ab - 2b^2 + 2a + 2b - 1$$
Simplifying further:
$$0 = -a^2 - 2ab - b^2 + 2a + 2b - 1$$
$$a^2 + 2ab + b^2 = 2a + 2b - 1$$

7. Since $a, b$ are positive integers, we need to find values such that $a + b < 3$. The only possible values are $a = 1$ and $b = 1$, which gives:
$$c = 1 + 1 - 1 = 1$$

8. Verifying the solution, we have $a = 1$, $b = 1$, and $c = 1$:
$$ab + ac + bc = 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 = 3$$
which is a prime number. Also:
$$\frac{1+1}{1+1} = \frac{2}{2} = 1 \quad \text{and} \quad \frac{1+1}{1+1} = \frac{2}{2} = 1$$
Thus, the given condition is satisfied.

The final answer is $\boxed{ (1, 1, 1) }$.