Find all pairs of prime numbers $p, q$ such that the number $pq + p - 6$ is also prime.
1. **Assume both \( p \) and \( q \) are odd primes:**

   If both \( p \) and \( q \) are odd primes, then \( pq + p - 6 \) is also odd. However, we need to check if it is prime. 

   Consider the expression modulo 2:
   \[
   pq + p - 6 \equiv 1 \cdot 1 + 1 - 6 \equiv 1 + 1 - 6 \equiv -4 \equiv 0 \pmod{2}
   \]
   This implies that \( pq + p - 6 \) is even. The only even prime number is 2. Therefore, we must have:
   \[
   pq + p - 6 = 2
   \]
   Rearranging, we get:
   \[
   p(q + 1) = 8
   \]
   Since \( p \) and \( q \) are odd primes, the product \( p(q + 1) = 8 \) has no solutions because 8 has no odd prime divisors.

2. **Consider the case where one of the primes is 2:**

   - **Case 1: \( p = 2 \)**

     Substitute \( p = 2 \) into the expression:
     \[
     2q + 2 - 6 = 2q - 4
     \]
     For \( 2q - 4 \) to be prime, it must be greater than 1. Therefore:
     \[
     2q - 4 > 1 \implies 2q > 5 \implies q > 2.5
     \]
     Since \( q \) is a prime number, the smallest possible value for \( q \) is 3. Substitute \( q = 3 \):
     \[
     2 \cdot 3 - 4 = 6 - 4 = 2
     \]
     2 is a prime number, so \( (p, q) = (2, 3) \) is a solution.

   - **Case 2: \( q = 2 \)**

     Substitute \( q = 2 \) into the expression:
     \[
     2p + p - 6 = 3p - 6
     \]
     For \( 3p - 6 \) to be prime, it must be greater than 1. Therefore:
     \[
     3p - 6 > 1 \implies 3p > 7 \implies p > \frac{7}{3} \approx 2.33
     \]
     Since \( p \) is a prime number, the smallest possible value for \( p \) is 3. Substitute \( p = 3 \):
     \[
     3 \cdot 3 - 6 = 9 - 6 = 3
     \]
     3 is a prime number, so \( (p, q) = (3, 2) \) is a solution.

Therefore, the only pairs of prime numbers \( (p, q) \) such that \( pq + p - 6 \) is also prime are \( (2, 3) \) and \( (3, 2) \).

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