Find all positive integers $x$, for which the equation
$$a+b+c=xabc$$ has solution in positive integers.
Solve the equation for these values of $x$
1. **Assume without loss of generality (WLOG) that \(a \leq b \leq c\).** This assumption simplifies the problem by reducing the number of cases we need to consider.

2. **Rewrite the given equation \(a + b + c = xabc\) and use the assumption \(a \leq b \leq c\):**
   \[
   a + b + c \leq 3c
   \]
   This implies:
   \[
   xabc \leq 3c \implies xab \leq 3
   \]
   Since \(a, b, c\) are positive integers, \(ab\) must be a divisor of 3. The possible pairs \((a, b)\) are \((1, 1)\), \((1, 2)\), and \((1, 3)\).

3. **Case 1: \((a, b) = (1, 1)\)**
   \[
   a + b + c = xabc \implies 1 + 1 + c = x \cdot 1 \cdot 1 \cdot c \implies 2 + c = xc
   \]
   Solving for \(c\):
   \[
   2 + c = xc \implies c(x - 1) = 2 \implies c = \frac{2}{x - 1}
   \]
   Since \(c\) must be a positive integer, \(x - 1\) must be a divisor of 2. The divisors of 2 are 1 and 2, giving \(x = 2 + 1\) and \(x = 2 + 2\):
   \[
   x - 1 = 1 \implies x = 2 \implies c = 2
   \]
   \[
   x - 1 = 2 \implies x = 3 \implies c = 1
   \]
   Thus, the solutions for this case are \((a, b, c, x) = (1, 1, 2, 2)\) and \((1, 1, 1, 3)\).

4. **Case 2: \((a, b) = (1, 2)\)**
   \[
   a + b + c = xabc \implies 1 + 2 + c = x \cdot 1 \cdot 2 \cdot c \implies 3 + c = 2xc
   \]
   Solving for \(c\):
   \[
   3 + c = 2xc \implies c(2x - 1) = 3 \implies c = \frac{3}{2x - 1}
   \]
   Since \(c\) must be a positive integer, \(2x - 1\) must be a divisor of 3. The divisors of 3 are 1 and 3, giving \(2x - 1 = 1\) and \(2x - 1 = 3\):
   \[
   2x - 1 = 1 \implies 2x = 2 \implies x = 1 \implies c = 3
   \]
   \[
   2x - 1 = 3 \implies 2x = 4 \implies x = 2 \implies c = 1.5 \quad (\text{not an integer})
   \]
   Thus, the solution for this case is \((a, b, c, x) = (1, 2, 3, 1)\).

5. **Case 3: \((a, b) = (1, 3)\)**
   \[
   a + b + c = xabc \implies 1 + 3 + c = x \cdot 1 \cdot 3 \cdot c \implies 4 + c = 3xc
   \]
   Solving for \(c\):
   \[
   4 + c = 3xc \implies c(3x - 1) = 4 \implies c = \frac{4}{3x - 1}
   \]
   Since \(c\) must be a positive integer, \(3x - 1\) must be a divisor of 4. The divisors of 4 are 1, 2, and 4, giving \(3x - 1 = 1\), \(3x - 1 = 2\), and \(3x - 1 = 4\):
   \[
   3x - 1 = 1 \implies 3x = 2 \implies x = \frac{2}{3} \quad (\text{not an integer})
   \]
   \[
   3x - 1 = 2 \implies 3x = 3 \implies x = 1 \implies c = 4
   \]
   \[
   3x - 1 = 4 \implies 3x = 5 \implies x = \frac{5}{3} \quad (\text{not an integer})
   \]
   Thus, the solution for this case is \((a, b, c, x) = (1, 3, 4, 1)\).

6. **Summary of solutions:**
   The solutions for \((a, b, c, x)\) are \((1, 1, 2, 2)\), \((1, 1, 1, 3)\), and \((1, 2, 3, 1)\). Permutations of \(a, b, c\) are also valid solutions.

The final answer is \( \boxed{ (1, 1, 2, 2) } \), \((1, 1, 1, 3)\), and \((1, 2, 3, 1)\).