If $$(2\alpha + 1,\; \alpha^2 - 3\alpha,\; \frac{\alpha - 1}{2})$$ is the image of $$(\alpha, 2\alpha, 1)$$ in the line $$\frac{x-2}{3} = \frac{y-1}{2} = \frac{z}{1}$$, then the possible value(s) of $$\alpha$$ is/are :

A

Only $$3$$

B

Only $$3$$ and $$-1$$

C

Only $$3\frac{1}{4}$$ and $$-1$$

D

Only $$3$$ and $$\frac{1}{4}$$