Determine all real numbers $ a$ such that the inequality $ |x^2 \plus{} 2ax \plus{} 3a|\le2$ has exactly one solution in $ x$.
To determine all real numbers \( a \) such that the inequality \( |x^2 + 2ax + 3a| \le 2 \) has exactly one solution in \( x \), we need to analyze the quadratic expression inside the absolute value and the conditions under which the inequality holds.

1. **Rewrite the inequality without the absolute value:**
   \[
   |x^2 + 2ax + 3a| \le 2
   \]
   This can be split into two inequalities:
   \[
   -2 \le x^2 + 2ax + 3a \le 2
   \]

2. **Analyze the quadratic expression:**
   The quadratic expression \( x^2 + 2ax + 3a \) is a parabola that opens upwards (since the coefficient of \( x^2 \) is positive). For the inequality to have exactly one solution, the vertex of the parabola must lie on the boundary of the inequality, i.e., the vertex must have a y-coordinate of either 2 or -2.

3. **Find the vertex of the parabola:**
   The vertex form of the quadratic expression \( x^2 + 2ax + 3a \) can be found by completing the square:
   \[
   x^2 + 2ax + 3a = (x + a)^2 - a^2 + 3a
   \]
   Therefore, the vertex of the parabola is at \( \left(-a, 3a - a^2\right) \).

4. **Set the y-coordinate of the vertex to 2:**
   For the inequality to have exactly one solution, the vertex must be on the boundary of the inequality:
   \[
   3a - a^2 = 2
   \]
   Solve this quadratic equation:
   \[
   a^2 - 3a + 2 = 0
   \]
   Factor the quadratic equation:
   \[
   (a - 2)(a - 1) = 0
   \]
   Therefore, the solutions are:
   \[
   a = 1 \quad \text{or} \quad a = 2
   \]

5. **Verify the solutions:**
   For \( a = 1 \):
   \[
   x^2 + 2(1)x + 3(1) = x^2 + 2x + 3
   \]
   The vertex is at \( (-1, 2) \), which satisfies the condition.

   For \( a = 2 \):
   \[
   x^2 + 2(2)x + 3(2) = x^2 + 4x + 6
   \]
   The vertex is at \( (-2, 2) \), which also satisfies the condition.

Thus, the values of \( a \) that satisfy the given condition are \( a = 1 \) and \( a = 2 \).

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