Find all sets $(a, b, c)$ of different positive integers $a$, $b$, $c$, for which: 

[b]*[/b] $2a - 1$ is a multiple of $b$;  

[b]*[/b] $2b - 1$ is a multiple of $c$; 

[b]*[/b] $2c - 1$ is a multiple of $a$.
To solve the problem, we need to find all sets of different positive integers \((a, b, c)\) such that:
1. \(2a - 1\) is a multiple of \(b\),
2. \(2b - 1\) is a multiple of \(c\),
3. \(2c - 1\) is a multiple of \(a\).

Let's analyze the conditions step by step.

1. **Assume \(a \geq b \geq c\)**:
   - This assumption helps us to simplify the problem, but we must remember to check all possible permutations of \((a, b, c)\) later.

2. **From the first condition \(2a - 1\) is a multiple of \(b\)**:
   - This implies \(2a - 1 = kb\) for some integer \(k\).
   - Rearranging, we get \(2a = kb + 1\).

3. **From the second condition \(2b - 1\) is a multiple of \(c\)**:
   - This implies \(2b - 1 = lc\) for some integer \(l\).
   - Rearranging, we get \(2b = lc + 1\).

4. **From the third condition \(2c - 1\) is a multiple of \(a\)**:
   - This implies \(2c - 1 = ma\) for some integer \(m\).
   - Rearranging, we get \(2c = ma + 1\).

Now, let's substitute and solve these equations step by step.

5. **Substitute \(a = 2c - 1\) into the first condition**:
   - From \(2a - 1 = kb\), we get \(2(2c - 1) - 1 = kb\).
   - Simplifying, \(4c - 3 = kb\).

6. **Substitute \(b = \frac{4c - 3}{k}\) into the second condition**:
   - From \(2b - 1 = lc\), we get \(2\left(\frac{4c - 3}{k}\right) - 1 = lc\).
   - Simplifying, \(\frac{8c - 6}{k} - 1 = lc\).
   - Further simplifying, \(\frac{8c - 6 - k}{k} = lc\).

7. **Solve for \(k\)**:
   - For \(b\) to be an integer, \(4c - 3\) must be divisible by \(k\).
   - Let's check possible values of \(k\).

8. **Check \(k = 1\)**:
   - If \(k = 1\), then \(b = 4c - 3\).
   - Substitute into the second condition: \(2(4c - 3) - 1 = lc\).
   - Simplifying, \(8c - 6 - 1 = lc\).
   - Further simplifying, \(8c - 7 = lc\).
   - This implies \(l = 8 - \frac{7}{c}\).
   - For \(l\) to be an integer, \(c\) must be 1 or 7.
   - If \(c = 1\), then \(a = 2(1) - 1 = 1\) and \(b = 4(1) - 3 = 1\), but \(a, b, c\) must be different.
   - If \(c = 7\), then \(a = 2(7) - 1 = 13\) and \(b = 4(7) - 3 = 25\).

9. **Verify the solution \((a, b, c) = (13, 25, 7)\)**:
   - Check the first condition: \(2a - 1 = 2(13) - 1 = 25\), which is a multiple of \(b = 25\).
   - Check the second condition: \(2b - 1 = 2(25) - 1 = 49\), which is a multiple of \(c = 7\).
   - Check the third condition: \(2c - 1 = 2(7) - 1 = 13\), which is a multiple of \(a = 13\).

Thus, the set \((a, b, c) = (13, 25, 7)\) satisfies all the conditions.

The final answer is \( \boxed{ (13, 25, 7) } \).