Find all triples of positive integers $ (p,q,n)$, with $ p$ and $ q$ primes, satisfying:

$ p(p\plus{}3)\plus{}q(q\plus{}3)\equal{}n(n\plus{}3)$.
1. We start with the given equation:
   \[
   p(p + 3) + q(q + 3) = n(n + 3)
   \]
   where \( p \) and \( q \) are primes, and \( n \) is a positive integer.

2. Consider the equation modulo 3. We have:
   \[
   p(p + 3) \equiv p^2 \pmod{3}
   \]
   and similarly,
   \[
   q(q + 3) \equiv q^2 \pmod{3}
   \]
   Therefore, the equation becomes:
   \[
   p^2 + q^2 \equiv n^2 \pmod{3}
   \]

3. Since \( p \) and \( q \) are primes, they can be either 2 or 3 modulo 3. If both \( p \) and \( q \) are not equal to 3, then they must be 2 modulo 3. However, \( 2^2 \equiv 1 \pmod{3} \), so:
   \[
   1 + 1 \equiv 2 \pmod{3}
   \]
   which cannot be equal to \( n^2 \pmod{3} \) since \( n^2 \equiv 0 \) or \( 1 \pmod{3} \). Therefore, one of \( p \) or \( q \) must be 3.

4. Without loss of generality, assume \( p = 3 \). Substituting \( p = 3 \) into the original equation, we get:
   \[
   3(3 + 3) + q(q + 3) = n(n + 3)
   \]
   Simplifying, we have:
   \[
   18 + q(q + 3) = n(n + 3)
   \]

5. Rearrange the equation to solve for \( n \):
   \[
   n^2 + 3n - q^2 - 3q - 18 = 0
   \]
   This is a quadratic equation in \( n \). The discriminant of this quadratic equation must be a perfect square for \( n \) to be an integer:
   \[
   \Delta = 9 + 4(q^2 + 3q + 18) = 4q^2 + 12q + 81
   \]
   Let \( \Delta = k^2 \) for some integer \( k \). Then:
   \[
   k^2 = 4q^2 + 12q + 81
   \]

6. Rearrange to form a difference of squares:
   \[
   k^2 - 81 = 4q^2 + 12q
   \]
   \[
   (k - 9)(k + 9) = 4q(q + 3)
   \]

7. We need to find integer solutions for \( k \) and \( q \). By trial and error, we find that \( k = 11 \) and \( q = 2 \) works:
   \[
   (11 - 9)(11 + 9) = 2 \cdot 5 \cdot 4 = 40
   \]
   \[
   2 \cdot 20 = 40
   \]

8. Substituting \( q = 2 \) back into the equation:
   \[
   18 + 2(2 + 3) = n(n + 3)
   \]
   \[
   18 + 10 = n(n + 3)
   \]
   \[
   28 = n(n + 3)
   \]

9. Solving the quadratic equation:
   \[
   n^2 + 3n - 28 = 0
   \]
   Using the quadratic formula:
   \[
   n = \frac{-3 \pm \sqrt{9 + 112}}{2} = \frac{-3 \pm 11}{2}
   \]
   \[
   n = 4 \quad \text{or} \quad n = -7
   \]
   Since \( n \) must be a positive integer, we have \( n = 4 \).

10. Therefore, the only solution is:
   \[
   (p, q, n) = (3, 2, 4)
   \]

The final answer is \(\boxed{(3, 2, 4)}\).