Let $a$ be a fixed integer. Find all integer solutions $x, y, z$ of the system
\[5x + (a + 2)y + (a + 2)z = a,\]\[(2a + 4)x + (a^2 + 3)y + (2a + 2)z = 3a - 1,\]\[(2a + 4)x + (2a + 2)y + (a^2 + 3)z = a + 1.\]
We are given the system of equations:
\[
5x + (a + 2)y + (a + 2)z = a,
\]
\[
(2a + 4)x + (a^2 + 3)y + (2a + 2)z = 3a - 1,
\]
\[
(2a + 4)x + (2a + 2)y + (a^2 + 3)z = a + 1.
\]

1. **Subtract the third equation from the second:**
\[
(2a + 4)x + (a^2 + 3)y + (2a + 2)z - \left[(2a + 4)x + (2a + 2)y + (a^2 + 3)z\right] = (3a - 1) - (a + 1).
\]
Simplifying, we get:
\[
(a^2 + 3)y + (2a + 2)z - (2a + 2)y - (a^2 + 3)z = 2a - 2.
\]
\[
(a^2 + 3 - 2a - 2)y + (2a + 2 - a^2 - 3)z = 2a - 2.
\]
\[
(a - 1)^2(y - z) = 2(a - 1).
\]

2. **Case 1: \(a = 1\)**
   - If \(a = 1\), the system becomes:
   \[
   5x + 3(y + z) = 1,
   \]
   \[
   6x + 4(y + z) = 2.
   \]
   - Subtract the first equation from the second:
   \[
   (6x + 4(y + z)) - (5x + 3(y + z)) = 2 - 1,
   \]
   \[
   x + (y + z) = 1.
   \]
   - Let \(y + z = k\), then \(x + k = 1\) and \(5x + 3k = 1\).
   - Solving for \(x\):
   \[
   x = 1 - k,
   \]
   \[
   5(1 - k) + 3k = 1,
   \]
   \[
   5 - 5k + 3k = 1,
   \]
   \[
   5 - 2k = 1,
   \]
   \[
   2k = 4,
   \]
   \[
   k = 2.
   \]
   - Therefore, \(x = 1 - 2 = -1\), and \(y + z = 2\).
   - The solutions are \((x, y, z) = (-1, n, 2 - n)\) for any integer \(n\).

3. **Case 2: \(a \neq 1\)**
   - From \((a - 1)^2(y - z) = 2(a - 1)\), we get:
   \[
   (a - 1)(a - 1)(y - z) = 2(a - 1).
   \]
   - If \(a \neq 1\), we can divide both sides by \(a - 1\):
   \[
   (a - 1)(y - z) = 2.
   \]
   - Therefore, \(y - z = \frac{2}{a - 1}\).
   - Since \(y - z\) is an integer, \(a - 1\) must be a divisor of 2. Thus, \(a - 1 \in \{\pm 1, \pm 2\}\), which gives \(a \in \{-1, 0, 2, 3\}\).

4. **Solving for each \(a\):**
   - **\(a = -1\):**
     \[
     5x + y + z = -1,
     \]
     \[
     2x + 4y + 0z = -4,
     \]
     \[
     2x + 0y + 4z = 0.
     \]
     - Solving, we get \((x, y, z) = (0, -1, 0)\).

   - **\(a = 0\):**
     \[
     5x + 2y + 2z = 0,
     \]
     \[
     4x + 3y + 2z = -1,
     \]
     \[
     4x + 2y + 3z = 1.
     \]
     - Solving, we get \((x, y, z) = (0, -1, 1)\).

   - **\(a = 2\):**
     \[
     5x + 4y + 4z = 2,
     \]
     \[
     8x + 7y + 6z = 5,
     \]
     \[
     8x + 6y + 7z = 3.
     \]
     - Solving, we get \((x, y, z) = (-6, 5, 3)\).

   - **\(a = 3\):**
     \[
     5x + 5y + 5z = 3,
     \]
     \[
     10x + 12y + 8z = 8,
     \]
     \[
     10x + 8y + 12z = 4.
     \]
     - This reduces to \(x + y + z = \frac{3}{5}\), which has no integer solutions.

Thus, all the cases are exhausted.

The final answer is \( \boxed{ (x, y, z) = (-1, n, 2 - n) } \) for \(a = 1\), \((x, y, z) = (0, -1, 0)\) for \(a = -1\), \((x, y, z) = (0, -1, 1)\) for \(a = 0\), and \((x, y, z) = (-6, 5, 3)\) for \(a = 2\). There are no solutions for \(a = 3\).