The system of equations can be written in the form $AX = B$, where $A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$. To examine the consistency, we first find the determinant of matrix $A$.

Expanding along the first row, we get: $|A| = 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a) = 4a - 2a - a = a$.

Case 1: If $a \neq 0$, then $|A| \neq 0$. In this case, the inverse of $A$ exists, and there exists a unique solution. Thus, the system is consistent.

Case 2: If $a = 0$, then $|A| = 0$. We calculate $(adj\ A)B$. For $a=0$, the matrix $A$ becomes $\begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 0 & 0 & 0 \end{bmatrix}$. The adjoint matrix is found to be $(adj\ A) = \begin{bmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

Now, $(adj\ A)B = \begin{bmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix} = \begin{bmatrix} -4 \\ 0 \\ 4 \end{bmatrix}$. Since $(adj\ A)B \neq 0$, there exists no solution. Hence, the system is inconsistent when $a = 0$.