Find all triples $(x, y, z)$ such that $x, y, z, x - y, y - z, x - z$ are all prime positive integers.
We need to find all triples \((x, y, z)\) such that \(x, y, z, x - y, y - z, x - z\) are all prime positive integers. Let's analyze the problem step by step.

1. **Distinct Primes**:
   Since \(x, y, z\) are distinct positive prime integers, the differences \(x - y, y - z, x - z\) must also be positive primes. Note that 2 is the only even prime number, so at most one of \(x, y, z\) can be even.

2. **Case 1: All Odd Primes**:
   If \(x, y, z\) are all odd primes, then \(x - y, y - z, x - z\) are even numbers. The only even prime is 2, so:
   \[
   x - y = 2, \quad y - z = 2 \implies x - z = 4
   \]
   However, 4 is not a prime number. Therefore, this case is not possible.

3. **Case 2: Two Odd Primes and One Even Prime**:
   Without loss of generality, assume \(z = 2\) (the only even prime). Then \(x\) and \(y\) are odd primes. We have:
   \[
   x - y = 2, \quad y - 2 = \text{prime}, \quad x - 2 = \text{prime}
   \]
   Since \(x = y + 2\), we need \(y - 2\) and \(y + 2 - 2 = y\) to be primes. Thus, \(y - 2, y, y + 2\) must all be primes.

   Let's consider the possible values of \(y\):
   - If \(y = 5\), then \(y - 2 = 3\) and \(y + 2 = 7\), which are all primes.
   - Therefore, \(x = y + 2 = 7\) and \(z = 2\).

   We check:
   \[
   x = 7, \quad y = 5, \quad z = 2
   \]
   \[
   x - y = 7 - 5 = 2, \quad y - z = 5 - 2 = 3, \quad x - z = 7 - 2 = 5
   \]
   All values are primes, so \((x, y, z) = (7, 5, 2)\) is a valid solution.

4. **Case 3: One Odd Prime and Two Even Primes**:
   Assume \(y = 2\) and \(x, z\) are odd primes. Then:
   \[
   x - 2 = \text{prime}, \quad 2 - z = \text{prime}, \quad x - z = \text{prime}
   \]
   Since \(2 - z\) must be positive, \(z\) must be 1, which is not a prime. Therefore, this case is not possible.

5. **Case 4: One Even Prime and Two Odd Primes**:
   Assume \(x = 2\) and \(y, z\) are odd primes. Then:
   \[
   2 - y = \text{prime}, \quad y - z = \text{prime}, \quad 2 - z = \text{prime}
   \]
   Since \(2 - y\) and \(2 - z\) must be positive, \(y\) and \(z\) must be 1, which is not a prime. Therefore, this case is not possible.

We have considered all cases, and the only solution is \(\boxed{(x, y, z) = (7, 5, 2)}\).