Find all integers x,y,z with $2\leq x\leq y\leq z$ st
$xy \equiv 1 $(mod z) $xz\equiv 1$(mod y) $yz \equiv 1$ (mod x)

To find all integers \( x, y, z \) such that \( 2 \leq x \leq y \leq z \) and the following congruences hold:
\[ xy \equiv 1 \pmod{z} \]
\[ xz \equiv 1 \pmod{y} \]
\[ yz \equiv 1 \pmod{x} \]

1. **Pairwise Coprime Condition:**
   Since \( xy \equiv 1 \pmod{z} \), \( xz \equiv 1 \pmod{y} \), and \( yz \equiv 1 \pmod{x} \), it implies that \( x, y, z \) must be pairwise coprime. This is because if any two of them shared a common factor, the modular inverses would not exist.

2. **Bounding \( x \):**
   Given \( 2 \leq x \leq y \leq z \), we start by bounding \( x \). We know:
   \[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = k + \frac{1}{xyz} \]
   for some integer \( k \). Since \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} > 1 \), we have:
   \[ 1 < \frac{1}{x} + \frac{1}{y} + \frac{1}{z} < \frac{3}{x} \]
   This implies:
   \[ x < 3 \]
   Given \( x \geq 2 \), we conclude:
   \[ x = 2 \]

3. **Bounding \( y \):**
   With \( x = 2 \), we have:
   \[ \frac{1}{2} + \frac{1}{y} + \frac{1}{z} > 1 \]
   Simplifying, we get:
   \[ \frac{1}{y} + \frac{1}{z} > \frac{1}{2} \]
   Since \( y \leq z \), we have:
   \[ \frac{1}{y} + \frac{1}{z} < \frac{2}{y} \]
   Thus:
   \[ \frac{1}{2} < \frac{2}{y} \]
   This implies:
   \[ y < 4 \]
   Given \( y \geq 2 \) and \( y \geq x \), we conclude:
   \[ y = 3 \]

4. **Bounding \( z \):**
   With \( x = 2 \) and \( y = 3 \), we have:
   \[ \frac{1}{2} + \frac{1}{3} + \frac{1}{z} > 1 \]
   Simplifying, we get:
   \[ \frac{5}{6} + \frac{1}{z} > 1 \]
   Thus:
   \[ \frac{1}{z} > \frac{1}{6} \]
   This implies:
   \[ z < 6 \]
   Given \( z \geq y \), we conclude:
   \[ z = 4 \text{ or } z = 5 \]

5. **Checking Pairwise Coprimality:**
   - For \( z = 4 \):
     \[ \gcd(2, 4) = 2 \]
     This violates the pairwise coprime condition.
   - For \( z = 5 \):
     \[ \gcd(2, 5) = 1 \]
     \[ \gcd(3, 5) = 1 \]
     \[ \gcd(2, 3) = 1 \]
     This satisfies the pairwise coprime condition.

6. **Verification:**
   - Check \( xy \equiv 1 \pmod{z} \):
     \[ 2 \cdot 3 = 6 \equiv 1 \pmod{5} \]
   - Check \( xz \equiv 1 \pmod{y} \):
     \[ 2 \cdot 5 = 10 \equiv 1 \pmod{3} \]
   - Check \( yz \equiv 1 \pmod{x} \):
     \[ 3 \cdot 5 = 15 \equiv 1 \pmod{2} \]

All conditions are satisfied for \( (x, y, z) = (2, 3, 5) \).

The final answer is \( \boxed{ (x, y, z) = (2, 3, 5) } \).