The correct answer is $\boxed{\text{D}}$.
A system of linear equations has infinitely many solutions if the augmented matrix has a nontrivial row of all zeros. In this case, the augmented matrix is
$$\left[ {\begin{array}{*{20}{c}} 2&1&{ – 4} & \alpha \ 4&3&{ – 12} & 5 \ 1&2&{ – 8} & 7 \end{array}} \right]$$
The second and third columns of this matrix are linearly dependent, so the matrix has a nontrivial row of all zeros. Therefore, the system of equations has infinitely many solutions.
To see this, let $x$ and $y$ be any two real numbers. Then we can solve the system of equations for $z$ as follows:
$$z = \frac{4x + 12y – 5\alpha}{1 – 2x}$$
Since $x$ and $y$ can be any real numbers, $z$ can also be any real number. Therefore, the system of equations has infinitely many solutions.