The correct answer is $\boxed{\text{B. either 0 or 1}}$.
To solve the system of equations, we can use Gaussian elimination. First, we can subtract the first equation from the second equation to get $x_2 + x_3 = 1$. Then, we can subtract $4$ times the first equation from the third equation to get $-3x_2 – 3ax_3 = -3$.
Now, we can express $x_2$ and $x_3$ in terms of $a$:
$$x_2 = \frac{1}{5} – a$$
$$x_3 = \frac{1}{5} – \frac{3a}{5}$$
Substituting these expressions into the first equation, we get:
$$x_1 + \frac{1}{5} – a + 2 \left( \frac{1}{5} – \frac{3a}{5} \right) = 1$$
Simplifying, we get:
$$x_1 – 2a = -\frac{4}{5}$$
Therefore, $x_1 = 2a – \frac{4}{5}$.
We know that the system of equations has a unique solution if the determinant of the coefficient matrix is not equal to zero. The coefficient matrix for this system of equations is:
$$\begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 1 & 4 & a \end{bmatrix}$$
The determinant of this matrix is:
$$\begin{vmatrix} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 1 & 4 & a \end{vmatrix} = -a^2 + 7a – 12$$
For the system of equations to have a unique solution, this determinant must not be equal to zero. Therefore, $a^2 – 7a + 12 \neq 0$.
Solving this equation, we get $a = 0$ or $a = 1$. Therefore, the only possible values for $a$ are $\boxed{\text{0 or 1}}$.