Since the second and third columns of the coefficient matrix are linearly deoendent, determis $0.$ So, the system of equations either has infinitely many solutions (if they are consistent) or no solution. To check for consistency, we apply reduction method on $(A\mid B)$
$R_{2}\leftarrow R_{2}-2R_{1},R_{3}\leftarrow R_{3}-0.5R_{1},R_{3}\leftarrow R_{3}-1.5R_{2}, R_1 \leftarrow R_1/2$
obtain the resultant matrix
$\large\begin{pmatrix}2&1&{-4}&\alpha\\4&3&{-12}&5\\1&2&{-8}&7\end{pmatrix}\rightarrow\begin{pmatrix}1&0.5&{-2}&0.5\alpha\\0&1&{-4}&5-2\alpha\\0&0&{-0}& -0.5+2.5\alpha\end{pmatrix}$
or infinitely many solutions, we must have $-0.5+2.5\alpha =0\; i.e., \alpha=\dfrac{1}{5}.$ So, for only $1$ value of $\alpha,$ this system has infinitely many solutions. So, option (B) is correct.