Find the greatest possible value of $pq + r$, where p, q, and r are (not necessarily distinct) prime numbers satisfying $pq + qr + rp = 2016$.

To find the greatest possible value of \( pq + r \), where \( p \), \( q \), and \( r \) are prime numbers satisfying \( pq + qr + rp = 2016 \), we can follow these steps:

1. **Minimize \( r \) and make \( p \) and \( q \) as close as possible**:
   - Start by setting \( r = 2 \) (the smallest prime number).

2. **Substitute \( r = 2 \) into the equation**:
   \[
   pq + 2q + 2p = 2016
   \]

3. **Rearrange the equation**:
   \[
   pq + 2q + 2p + 4 = 2020
   \]

4. **Factor the equation using Simon's Favorite Factoring Trick (SFFT)**:
   \[
   (p + 2)(q + 2) = 2020
   \]

5. **Find factor pairs of 2020**:
   - The factor pairs of 2020 are: \( (1, 2020), (2, 1010), (4, 505), (5, 404), (10, 202), (20, 101), (101, 20), (202, 10), (404, 5), (505, 4), (1010, 2), (2020, 1) \).

6. **Identify valid prime pairs**:
   - We need \( p \) and \( q \) to be prime numbers. Therefore, we check the factor pairs to see if \( p = a - 2 \) and \( q = b - 2 \) are both prime numbers.

7. **Check the factor pair \( (4, 505) \)**:
   - For \( (p + 2, q + 2) = (4, 505) \):
     \[
     p = 4 - 2 = 2 \quad \text{and} \quad q = 505 - 2 = 503
     \]
   - Both 2 and 503 are prime numbers.

8. **Calculate \( pq + r \)**:
   \[
   pq + r = 2 \cdot 503 + 2 = 1006 + 2 = 1008
   \]

9. **Verify that no other factor pairs yield a higher value**:
   - Since \( r \) must be minimized and any other prime \( r \) would be odd, making \( 2016 + r^2 \) odd, the factors of \( 2016 + r^2 \) would also be odd. This would not yield valid prime pairs for \( p \) and \( q \).

Therefore, the greatest possible value of \( pq + r \) is \( \boxed{1008} \).