The correct answer is $\boxed{\text{D}}$. There is no such value of $a$ for which the system of equations has a solution.
To see this, we can express the system of equations in matrix form as follows:
$$\begin{bmatrix}
2 & 3 & 0 \\
1 & 1 & 1 \\
1 & 2 & -1
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix} =
\begin{bmatrix}
4 \\
4 \\
a
\end{bmatrix}$$
The determinant of the coefficient matrix is $-1$. This means that the system of equations is inconsistent and has no solution.
We can also see this by considering the third equation. If we substitute the first two equations into the third equation, we get the equation $x + 2y – (2x + 3y) = a$. This simplifies to $y = a – 4$. Substituting this into the first equation, we get the equation $2x + 3(a – 4) = 4$. This simplifies to $2x – 6 = 0$. This equation has no real solutions, so there is no value of $a$ for which the system of equations has a solution.