Determine all positive integers $a,b,c$ such that $ab + ac + bc$ is a prime number and
$$\frac{a+b}{a+c}=\frac{b+c}{b+a}.$$
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 \( \boxed{ (1, 1, 1) } \).