Find all prime numbers $p$ for which there exist positive integers $x$, $y$, and $z$ such that the number

$x^p + y^p + z^p - x - y - z$

is a product of exactly three distinct prime numbers. 
To find all prime numbers \( p \) for which there exist positive integers \( x \), \( y \), and \( z \) such that the number \( x^p + y^p + z^p - x - y - z \) is a product of exactly three distinct prime numbers, we proceed as follows:

1. **Verification for \( p = 2, 3, 5 \)**:
   - For \( p = 2 \):
     \[
     x^2 + y^2 + z^2 - x - y - z
     \]
     Choose \( (x, y, z) = (1, 1, 6) \):
     \[
     1^2 + 1^2 + 6^2 - 1 - 1 - 6 = 1 + 1 + 36 - 1 - 1 - 6 = 30
     \]
     \( 30 = 2 \times 3 \times 5 \), which is a product of three distinct primes.
   
   - For \( p = 3 \):
     \[
     x^3 + y^3 + z^3 - x - y - z
     \]
     Choose \( (x, y, z) = (1, 2, 3) \):
     \[
     1^3 + 2^3 + 3^3 - 1 - 2 - 3 = 1 + 8 + 27 - 1 - 2 - 3 = 30
     \]
     \( 30 = 2 \times 3 \times 5 \), which is a product of three distinct primes.
   
   - For \( p = 5 \):
     \[
     x^5 + y^5 + z^5 - x - y - z
     \]
     Choose \( (x, y, z) = (1, 1, 2) \):
     \[
     1^5 + 1^5 + 2^5 - 1 - 1 - 2 = 1 + 1 + 32 - 1 - 1 - 2 = 30
     \]
     \( 30 = 2 \times 3 \times 5 \), which is a product of three distinct primes.

2. **Proof of necessity for \( p > 5 \)**:
   - Suppose \( p > 5 \). Note that by Fermat's Little Theorem, for any integer \( a \):
     \[
     a^p \equiv a \pmod{p}
     \]
     Therefore:
     \[
     x^p \equiv x \pmod{p}, \quad y^p \equiv y \pmod{p}, \quad z^p \equiv z \pmod{p}
     \]
     This implies:
     \[
     x^p + y^p + z^p - x - y - z \equiv 0 \pmod{p}
     \]
     Hence, \( p \mid (x^p + y^p + z^p - x - y - z) \).

   - Additionally, for any odd prime \( p \) and positive integer \( a \):
     \[
     a^p \equiv a \pmod{2} \quad \text{and} \quad a^p \equiv a \pmod{3}
     \]
     Proof:
     - If \( 2 \mid a \), then \( a^p \) and \( a \) are both even.
     - If \( 2 \nmid a \), then \( a^p \) and \( a \) are both odd.
     - If \( 3 \mid a \), then \( 3 \mid a^p \).
     - If \( 3 \nmid a \), then by Fermat's Little Theorem, \( a^{p-1} \equiv 1 \pmod{3} \), so \( a^p \equiv a \pmod{3} \).

   - Thus:
     \[
     6 \mid (x^p + y^p + z^p - x - y - z)
     \]
     This implies:
     \[
     x^p + y^p + z^p - x - y - z = 6k \quad \text{for some integer } k
     \]
     Since \( x^p + y^p + z^p - x - y - z \) is a product of exactly three distinct primes, let \( x^p + y^p + z^p - x - y - z = p_1 p_2 p_3 \) where \( p_1, p_2, p_3 \) are distinct primes.

   - For \( p > 7 \), consider:
     \[
     x^p + y^p + z^p - x - y - z = (x^p - x) + (y^p - y) + (z^p - z)
     \]
     Since at least one of \( x, y, z \) is greater than 1, we have:
     \[
     (x^p - x) + (y^p - y) + (z^p - z) \ge 2^p - 2
     \]
     For \( p > 7 \):
     \[
     2^p - 2 > 6p
     \]
     This leads to a contradiction because \( x^p + y^p + z^p - x - y - z \) cannot be a product of exactly three distinct primes if it exceeds \( 6p \).

Therefore, the only possible primes \( p \) are \( 2, 3, \) and \( 5 \).

The final answer is \( \boxed{2, 3, 5} \).