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 :

Let the original point be $$P = (\alpha, 2\alpha, 1)$$ and its image point be $$Q = \left(2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2}\right)$$.

$$\frac{x-2}{3} = \frac{y-1}{2} = \frac{z}{1}$$

The direction vector of this line is $$\vec{d} = 3\hat{i} + 2\hat{j} + 1\hat{k}$$.

$$\vec{PQ} = Q - P = \left((2\alpha + 1) - \alpha, \ (\alpha^2 - 3\alpha) - 2\alpha, \ \frac{\alpha - 1}{2} - 1\right)$$

$$\vec{PQ} = \left(\alpha + 1, \ \alpha^2 - 5\alpha, \ \frac{\alpha - 3}{2}\right)$$

$$\vec{PQ} \cdot \vec{d} = 0$$ (perpendicular to each other)

$$3(\alpha + 1) + 2(\alpha^2 - 5\alpha) + 1\left(\frac{\alpha - 3}{2}\right) = 0$$

$$4\alpha^2 - 13\alpha + 3 = 0$$

$$\alpha = 3 \quad \text{or} \quad \alpha = \frac{1}{4}$$

The midpoint $$M$$ of $$PQ$$ is given by $$\left(\frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}, \frac{z_P + z_Q}{2}\right)$$:

$$M = \left(\frac{\alpha + (2\alpha + 1)}{2}, \ \frac{2\alpha + (\alpha^2 - 3\alpha)}{2}, \ \frac{1 + \frac{\alpha - 1}{2}}{2}\right) = \left(\frac{3\alpha + 1}{2}, \ \frac{\alpha^2 - \alpha}{2}, \ \frac{\alpha + 1}{4}\right)$$

Substitute $$\alpha = 3$$ into the midpoint formula: $$M = \left(\frac{3(3) + 1}{2}, \ \frac{3^2 - 3}{2}, \ \frac{3 + 1}{4}\right) = (5, 3, 1)$$

Now, substitute $$(5, 3, 1)$$ into the line ratios: $$\frac{5-2}{3} = \frac{3-1}{2} = \frac{1}{1} \implies 1 = 1 = 1 \quad \text{(True)}$$

Therefore, $$\alpha = 3$$ is a valid solution.

However, $$\alpha = \frac{1}{4}$$ does not give equal ratios, and hence, an invalid solution.

$$M = \left(\frac{3(\frac{1}{4}) + 1}{2}, \ \frac{(\frac{1}{4})^2 - \frac{1}{4}}{2}, \ \frac{\frac{1}{4} + 1}{4}\right) = \left(\frac{7}{8}, \ -\frac{3}{32}, \ \frac{5}{16}\right)$$

$$\text{First ratio: } \frac{\frac{7}{8} - 2}{3} = \frac{-\frac{9}{8}}{3} = -\frac{3}{8}$$

$$\text{Third ratio: } \frac{\frac{5}{16}}{1} = \frac{5}{16}$$