. Find all triplets $(x, y, z)$, $x > y > z$ of positive integers such that $\frac{1}{x}+\frac{2}{y}+\frac{3}{z}= 1$
To find all triplets \((x, y, z)\) of positive integers such that \(x > y > z\) and \(\frac{1}{x} + \frac{2}{y} + \frac{3}{z} = 1\), we start by analyzing the given equation and the constraints.

1. **Initial Inequality Analysis:**
   \[
   1 = \frac{1}{x} + \frac{2}{y} + \frac{3}{z} < \frac{1}{z} + \frac{2}{z} + \frac{3}{z} = \frac{6}{z}
   \]
   This implies:
   \[
   \frac{6}{z} > 1 \implies z < 6
   \]
   Since \(z\) is a positive integer, we have:
   \[
   z \leq 5
   \]

2. **Lower Bound for \(z\):**
   Since \(\frac{1}{x} + \frac{2}{y} > 0\), we must have:
   \[
   1 - \frac{3}{z} > 0 \implies \frac{3}{z} < 1 \implies z > 3
   \]
   Combining the bounds, we get:
   \[
   3 < z < 6 \implies z = 4 \text{ or } z = 5
   \]

3. **Case 1: \(z = 4\):**
   \[
   \frac{1}{x} + \frac{2}{y} = 1 - \frac{3}{4} = \frac{1}{4}
   \]
   \[
   \frac{1}{4} = \frac{1}{x} + \frac{2}{y} < \frac{3}{y} \implies y < 12
   \]
   Also:
   \[
   \frac{1}{x} = \frac{1}{4} - \frac{2}{y} = \frac{y - 8}{4y} \implies y > 8
   \]
   Therefore:
   \[
   8 < y < 12 \implies y = 9 \text{ or } y = 10
   \]

   - For \(y = 9\):
     \[
     \frac{1}{x} = \frac{1}{4} - \frac{2}{9} = \frac{9 - 8}{36} = \frac{1}{36} \implies x = 36
     \]
     Thus, one solution is \((x, y, z) = (36, 9, 4)\).

   - For \(y = 10\):
     \[
     \frac{1}{x} = \frac{1}{4} - \frac{2}{10} = \frac{5 - 4}{20} = \frac{1}{20} \implies x = 20
     \]
     Thus, another solution is \((x, y, z) = (20, 10, 4)\).

4. **Case 2: \(z = 5\):**
   \[
   \frac{1}{x} + \frac{2}{y} = 1 - \frac{3}{5} = \frac{2}{5}
   \]
   \[
   \frac{2}{5} = \frac{1}{x} + \frac{2}{y} < \frac{3}{y} \implies y < 7.5
   \]
   Since \(y\) is an integer:
   \[
   y \leq 7
   \]
   Also:
   \[
   \frac{1}{x} = \frac{2}{5} - \frac{2}{y} = \frac{2(y - 5)}{5y} \implies y > 5
   \]
   Therefore:
   \[
   5 < y \leq 7 \implies y = 6 \text{ or } y = 7
   \]

   - For \(y = 6\):
     \[
     \frac{1}{x} = \frac{2}{5} - \frac{2}{6} = \frac{12 - 10}{30} = \frac{1}{15} \implies x = 15
     \]
     Thus, another solution is \((x, y, z) = (15, 6, 5)\).

   - For \(y = 7\):
     \[
     \frac{1}{x} = \frac{2}{5} - \frac{2}{7} = \frac{14 - 10}{35} = \frac{4}{35} \implies x = 8.75
     \]
     Since \(x\) must be an integer, \(y = 7\) does not provide a valid solution.

Therefore, the valid solutions are:
\[
(x, y, z) = (36, 9, 4), (20, 10, 4), (15, 6, 5)
\]

The final answer is \( \boxed{ (x, y, z) = \{(36, 9, 4), (20, 10, 4), (15, 6, 5)\} } \)