Determine the prime numbers $p, q, r$ with the property that: $p(p-7) + q (q-7) = r (r-7)$.
We are given the equation \( p(p-7) + q(q-7) = r(r-7) \) and need to find the prime numbers \( p, q, r \) that satisfy this equation. We will consider different cases based on the values of \( p, q, \) and \( r \).

1. **Assume \( p \geq q \):**
   - If \( p, q, r > 7 \), then \( r(r-7) > p(p-7) \implies r > p \).
   - Rewrite the equation as:
     \[
     p(p-7) + q(q-7) = r(r-7)
     \]
     \[
     p(p-7) = r^2 - 7r - q^2 + 7q
     \]
     \[
     p(p-7) = (r-q)(r+q-7)
     \]
     This implies \( p \mid (r-q)(r+q-7) \).

2. **Consider the divisibility conditions:**
   - If \( p \mid r-q \), then \( r-q \geq p \). This leads to:
     \[
     r+q-7 \leq p-7 < p \leq r-q \implies 2q < 7
     \]
     This is a contradiction since \( q \) is a prime number greater than 2.
   - Therefore, \( p \mid r+q-7 \implies r+q = kp + 7 \) for some \( k \in \mathbb{N} \).

3. **Analyze the possible values of \( k \):**
   - If \( k \geq 3 \), then \( r \geq 2p + 7 \). This leads to:
     \[
     r(r-7) \leq 2p(p-7)
     \]
     \[
     (2p+7)(2p) \leq 2p(p-7) \implies 4p^2 + 14p \leq 2p^2 - 14p
     \]
     \[
     4p^2 + 14p \leq 2p^2 - 14p \implies 2p^2 + 28p \leq 0
     \]
     This is a contradiction since \( p \) is a positive prime number.
   - Therefore, \( k = 0, 1, 2 \).

4. **Consider each value of \( k \):**
   - \( k = 0 \):
     \[
     r + q = 7
     \]
     This is a contradiction since \( p \geq q \geq 7 \).
   - \( k = 1 \):
     \[
     r + q = p + 7 < r + 7 \implies q < 7
     \]
     This is a contradiction since \( q \) is a prime number greater than 2.
   - \( k = 2 \):
     \[
     r + q = 2p + 7
     \]
     \[
     p(p-7) = (r-2p-7 + r)2p \implies 5p + 7 = 4r
     \]
     \[
     -p + 1 \equiv r \pmod{3} \implies p, r \equiv 2 \pmod{3}
     \]
     Therefore, \( q = 2p + 7 - r \equiv 4 + 7 - 2 \equiv 0 \pmod{3} \), which is a contradiction since \( q \) is a prime number.

5. **Consider \( r = 2, 3, 5, 7 \):**
   - For \( r = 7 \):
     \[
     p^2 - 7p - r^2 + 7r = -q(q-7) \implies (p-r)(p+r-7) = q(7-q)
     \]
     - If \( q = 7 \):
       \[
       p = r \text{ or } p + r = 7 \implies (p, q, r) = (a, 7, a) \text{ or } (5, 2, 7)
       \]
     - If \( q = 5 \):
       \[
       (p-r)(p+r-7) = 10 \implies (p, q, r) = (7, 5, 5) \text{ or } (7, 5, 2)
       \]
     - If \( q = 3 \):
       \[
       (p-r)(p+r-7) = 21 \implies (p, q, r) = (7, 3, 3)
       \]
     - If \( q = 2 \):
       \[
       (p-r)(p+r-7) = 10 \implies (p, q, r) = (7, 5, 5) \text{ or } (7, 5, 2)
       \]

6. **Summarize the solutions:**
   The solutions are:
   \[
   (p, q, r) = (2, 5, 7), (2, 5, 7), (7, 5, 5), (5, 7, 5), (5, 7, 2), (7, 5, 2), (7, 3, 3), (3, 7, 3), (a, 7, a), (7, a, a) \text{ where } a \text{ is prime}
   \]

The final answer is \( \boxed{ (p, q, r) = (2, 5, 7), (2, 5, 7), (7, 5, 5), (5, 7, 5), (5, 7, 2), (7, 5, 2), (7, 3, 3), (3, 7, 3), (a, 7, a), (7, a, a) \text{ where } a \text{ is prime} } \)