Find all triplets $(a, b, c)$ of positive integers, such that $a+bc, b+ac, c+ab$ are primes and all divide $(a^2+1)(b^2+1)(c^2+1)$.
1. We start by noting that \(a + bc\), \(b + ac\), and \(c + ab\) are primes and all divide \((a^2 + 1)(b^2 + 1)(c^2 + 1)\). Let's denote these primes by \(p_1\), \(p_2\), and \(p_3\) respectively.

2. If these three primes are distinct, their product would divide \((a^2 + 1)(b^2 + 1)(c^2 + 1)\). Therefore, we have:
   \[
   (a + bc)(b + ac)(c + ab) \leq (a^2 + 1)(b^2 + 1)(c^2 + 1)
   \]

3. Expanding the left-hand side:
   \[
   (a + bc)(b + ac)(c + ab) = abc(a + b + c) + a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + abc + a + b + c
   \]

4. Expanding the right-hand side:
   \[
   (a^2 + 1)(b^2 + 1)(c^2 + 1) = a^2b^2c^2 + a^2b^2 + a^2c^2 + b^2c^2 + a^2 + b^2 + c^2 + 1
   \]

5. For the inequality to hold, we need:
   \[
   abc(a + b + c) + a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + abc + a + b + c \leq a^2b^2c^2 + a^2b^2 + a^2c^2 + b^2c^2 + a^2 + b^2 + c^2 + 1
   \]

6. Simplifying, we get:
   \[
   (abc - 1)(a^2 + b^2 + c^2 + 1) \leq 0
   \]

7. Since \(a\), \(b\), and \(c\) are positive integers, \(abc - 1 \geq 0\) and \(a^2 + b^2 + c^2 + 1 > 0\). Therefore, the inequality \((abc - 1)(a^2 + b^2 + c^2 + 1) \leq 0\) is clearly wrong unless \(abc = 1\).

8. Hence, two of the primes must be the same. Without loss of generality, assume \(a + bc = b + ac\). This implies:
   \[
   (a - b)(c - 1) = 0
   \]

9. Therefore, either \(a = b\) or \(c = 1\).

10. If \(a = b\), then \(a + bc = a(c + 1)\) is prime. This implies \(a = 1\) and \(c + 1\) is prime. Since \(c + 1\) divides \(4(c^2 + 1)\), it must divide 8. Hence, \(c = 1\) and we get the solution \((1, 1, 1)\).

11. If \(c = 1\), then \(a + b\) and \(a + b\) are primes. This implies \(a = 1\) and \(b = 1\), leading to the same solution \((1, 1, 1)\).

12. If all three of \(a, b, c\) are distinct, we get that \(ab + 1\) and \(a + b\) are primes \(\geq 3\) dividing \((a^2 + 1)(b^2 + 1)\). But if \(a < b\), then \(ab + 1 > a^2 + 1\) and hence \(ab + 1 \mid b^2 + 1\). This implies \(ab + 1 \mid b^2 - ab = b(b - a)\), but both factors are positive and smaller than \(ab + 1\), leading to a contradiction.

13. Therefore, the only solution is \((1, 1, 1)\).

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