$A.X = B \\ \implies \begin{bmatrix} 1 &1 &2 \\ 1&2 &3 \\ 1& 4 & \alpha \end{bmatrix} \begin{bmatrix} x\\y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\2 \\4 \end{bmatrix}$
So, $X = A^{-1} B$
$A^{-1} =\dfrac{adj(A)}{det(A)}$
$\because \left[adj(A) = Transporse (cofactor(A)), cofactor(A) = (-1)^{i+j} minor(A) \right]$
$adj(A) = \begin{bmatrix} 2\alpha -12 &8-\alpha &-1 \\ 3 - \alpha&\alpha - 2 &-1 \\ 2& -3 & 1 \end{bmatrix} $
$det(A) = 2\alpha - 12 + 3 - \alpha + 4 = \alpha - 5$
Now, $A^{-1}B = \frac{1}{\alpha - 5} \begin{bmatrix} 0\\ \alpha -5 \\ 0 \end{bmatrix} = \begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix}$
So, the solution is independent of $\alpha$ and any real value of $\alpha$ would suffice. (This can be seen even by observing the equations carefully- each equation value is dependent only on $x_2$). But a value of 5 would cause the determinant to be 0 and multiple solutions to the given equation. So, any real value except 5 should be the answer- none of the choices is correct.
Ref: http://www.mathwords.com/i/inverse_of_a_matrix.htm