Determine all real numbers $a$ for which the system of equations
\begin{align*}
3x^2+2y^2+2z^2&=a\\
4x^2+4y^2+5z^2&=1-a
\end{align*}
has at least one solution $(x,y,z)$ in the real numbers.
1. **Substitute variables**:
   We start by substituting \( u = x^2 \), \( v = y^2 \), and \( w = z^2 \). This transforms the system of equations into:
   \[
   3u + 2v + 2w = a
   \]
   \[
   4u + 4v + 5w = 1 - a
   \]

2. **Combine equations**:
   To eliminate \( a \), we add the two equations:
   \[
   (3u + 2v + 2w) + (4u + 4v + 5w) = a + (1 - a)
   \]
   Simplifying, we get:
   \[
   7u + 6v + 7w = 1
   \]

3. **Determine the range of \( a \)**:
   We need to find the possible values of \( 3u + 2v + 2w \) given that \( 7u + 6v + 7w = 1 \) and \( u, v, w \geq 0 \). This is a linear function over a triangular domain defined by the constraints.

4. **Vertices of the triangular domain**:
   The vertices of the triangular domain can be found by setting two of the variables to zero and solving for the third:
   - For \( u \):
     \[
     7u = 1 \implies u = \frac{1}{7}, \quad v = 0, \quad w = 0
     \]
   - For \( v \):
     \[
     6v = 1 \implies v = \frac{1}{6}, \quad u = 0, \quad w = 0
     \]
   - For \( w \):
     \[
     7w = 1 \implies w = \frac{1}{7}, \quad u = 0, \quad v = 0
     \]

5. **Evaluate \( 3u + 2v + 2w \) at the vertices**:
   - At \( \left( \frac{1}{7}, 0, 0 \right) \):
     \[
     3u + 2v + 2w = 3 \left( \frac{1}{7} \right) + 2(0) + 2(0) = \frac{3}{7}
     \]
   - At \( \left( 0, \frac{1}{6}, 0 \right) \):
     \[
     3u + 2v + 2w = 3(0) + 2 \left( \frac{1}{6} \right) + 2(0) = \frac{2}{6} = \frac{1}{3}
     \]
   - At \( \left( 0, 0, \frac{1}{7} \right) \):
     \[
     3u + 2v + 2w = 3(0) + 2(0) + 2 \left( \frac{1}{7} \right) = \frac{2}{7}
     \]

6. **Determine the range of \( a \)**:
   Since \( 3u + 2v + 2w \) must lie between the minimum and maximum values obtained at the vertices, \( a \) must belong to the interval:
   \[
   \left[ \frac{2}{7}, \frac{3}{7} \right]
   \]

The final answer is \( \boxed{ \left[ \frac{2}{7}, \frac{3}{7} \right] } \)