Find all solutions in positive integers to:
\begin{eqnarray*} a + b + c = xyz \\ x + y + z = abc \end{eqnarray*}
We are given the system of equations:
\[
\begin{cases}
a + b + c = xyz \\
x + y + z = abc
\end{cases}
\]
We need to find all solutions in positive integers.

1. **Initial Analysis:**
   We start by analyzing the given equations:
   \[
   a + b + c = xyz
   \]
   \[
   x + y + z = abc
   \]

2. **Considering the Case \(a, b, c > 1\):**
   If \(a, b, c > 1\), then \((a-1)(b-1) > 0\). This implies:
   \[
   (a-1)(b-1) \geq 1 \implies ab - a - b + 1 \geq 1 \implies ab \geq a + b
   \]
   Similarly, for \(c > 1\), we have:
   \[
   abc \geq a + b + c
   \]

3. **Considering the Case \(x, y, z > 1\):**
   If \(x, y, z > 1\), then:
   \[
   xyz - x - y - z \geq 1
   \]
   This implies:
   \[
   xyz \geq x + y + z + 1
   \]

4. **Case Analysis:**
   - **Case 1: \(x, y, z > 1\):**
     If \(x, y, z > 1\), then:
     \[
     abc - a - b - c \geq 1 \quad \text{and} \quad xyz - x - y - z \geq 1
     \]
     This implies:
     \[
     abc \geq a + b + c + 1 \quad \text{and} \quad xyz \geq x + y + z + 1
     \]
     However, this contradicts the given equations \(a + b + c = xyz\) and \(x + y + z = abc\).

   - **Case 2: At least one of \(x, y, z\) is 1:**
     Assume \(z = 1\). Then the equations become:
     \[
     a + b + c = xy \quad \text{and} \quad x + y + 1 = abc
     \]
     We need to find \(x, y, a, b, c\) such that these equations hold.

5. **Finding Solutions:**
   - Assume \(z = 1\) and \(c = 1\):
     \[
     a + b + 1 = xy \quad \text{and} \quad x + y + 1 = ab
     \]
     We need to solve these equations for positive integers \(a, b, x, y\).

     - **Subcase 1: \(x = 2, y = 2\):**
       \[
       a + b + 1 = 2 \cdot 2 = 4 \implies a + b = 3
       \]
       \[
       2 + 2 + 1 = ab \implies ab = 5
       \]
       The only positive integer solutions for \(a\) and \(b\) that satisfy both equations are \(a = 1, b = 2\) or \(a = 2, b = 1\).

     - **Subcase 2: \(x = 3, y = 2\):**
       \[
       a + b + 1 = 3 \cdot 2 = 6 \implies a + b = 5
       \]
       \[
       3 + 2 + 1 = ab \implies ab = 6
       \]
       The only positive integer solutions for \(a\) and \(b\) that satisfy both equations are \(a = 2, b = 3\) or \(a = 3, b = 2\).

     - **Subcase 3: \(x = 3, y = 3\):**
       \[
       a + b + 1 = 3 \cdot 3 = 9 \implies a + b = 8
       \]
       \[
       3 + 3 + 1 = ab \implies ab = 7
       \]
       There are no positive integer solutions for \(a\) and \(b\) that satisfy both equations.

6. **Conclusion:**
   The only possible solutions in positive integers are:
   \[
   (x, y, z) = (3, 2, 1) \quad \text{and} \quad (a, b, c) = (3, 2, 1)
   \]
   \[
   (x, y, z) = (3, 3, 1) \quad \text{and} \quad (a, b, c) = (5, 2, 1)
   \]
   \[
   (x, y, z) = (5, 2, 1) \quad \text{and} \quad (a, b, c) = (3, 3, 1)
   \]

The final answer is \( \boxed{ (3, 2, 1), (3, 2, 1) } \), \((3, 3, 1), (5, 2, 1)\), \((5, 2, 1), (3, 3, 1)\).