Let $A = x^p + y^p + z^p - x - y - z$. For $p = 2$, we take $x = y = 4$ and $z = 3$. Then $A = 30 = 2 \cdot 3 \cdot 5$. For $p = 3$ we can take $x = 3$ and $y = 2$ and $z = 1$. Then again $A = 30 = 2 \cdot 3 \cdot 5$. For $p = 5$ we can take $x = 2$ and $y = 1$ and $z = 1$. Again $A = 30 = 2 \cdot 3 \cdot 5$.

Assume now that $p \geq 7$. Working modulo 2 and modulo 3 we see that $A$ is divisible by both 2 and 3. Moreover, by Fermat's Little Theorem, we have

$$x^p + y^p + z^p - x - y - z \equiv x + y + z - x - y - z = 0 \bmod p.$$

Therefore, by the given condition, we have to solve the equation

$$x^p + y^p + z^p - x - y - z = 6p.$$

If one of the numbers $x, y$ and $z$ is bigger than or equal to 2, let's say $x \geq 2$, then

$$6p \geq x^p - x = x(x^{p-1} - 1) \geq 2(2^{p-1} - 1) = 2^p - 2.$$

It is easy to check by induction that $2^n - 2 > 6n$ for all natural numbers $n \geq 6$. This contradiction shows that there are no more values of $p$ which satisfy the required property.

Remark. There are a couple of other ways to prove that $2^p - 2 > 6p$ for $p \geq 7$. For example, we can use the Binomial Theorem as follows:

$$2^p - 2 \geq 1 + p + \frac{p(p-1)}{2} + \frac{p(p-1)(p-2)}{6} - 2 \geq 1 + p + 3p + 5p - 2 > 6p.$$

We can also use Bernoulli's Inequality as follows:

$$2^p - 2 = 8(1+1)^{p-3} - 2 \geq 8(1 + (p-3)) - 2 = 8p - 18 > 6p.$$

The last inequality is true for $p \geq 11$. For $p = 7$ we can see directly that $2^p - 2 > 6p$.

One can also use calculus to show that $f(x) = 2^x - 6x$ is increasing for $x \geq 5$.

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 $\boxed{2, 3, 5}$.