Let $$A$$, $$B$$, $$C$$ be three points whose position vectors respectively are:
$$\vec{a} = \hat{i} + 4\hat{j} + 3\hat{k}$$
$$\vec{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$$, $$\alpha \in R$$
$$\vec{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$$
If $$\alpha$$ is the smallest positive integer for which $$\vec{a}$$, $$\vec{b}$$, $$\vec{c}$$ are non-collinear, then the length of the median, through $$A$$, of $$\triangle ABC$$ is:

We have the vectors $$\vec{a} = \hat{i} + 4\hat{j} + 3\hat{k}$$, $$\vec{b} = 2\hat{i} + \alpha\hat{j} + 4\hat{k}$$ and $$\vec{c} = 3\hat{i} - 2\hat{j} + 5\hat{k}$$. Computing $$\vec{AB} = \vec{b} - \vec{a} = \hat{i} + (\alpha-4)\hat{j} + \hat{k}$$ and $$\vec{AC} = \vec{c} - \vec{a} = 2\hat{i} - 6\hat{j} + 2\hat{k}$$, we set $$\vec{AB} = \lambda\vec{AC}$$. From $$1 = 2\lambda$$ we get $$\lambda = 1/2$$, and $$\alpha - 4 = -6\times\frac12 = -3$$ gives $$\alpha = 1$$. Verifying $$1 = 2\times\frac12 = 1$$, so the points are collinear when $$\alpha = 1$$.

Since $$\alpha = 1$$ makes them collinear, the smallest positive integer for non-collinearity is $$\alpha = 2$$.

With $$\alpha = 2$$, the points are $$B = (2,2,4)$$ and $$C = (3,-2,5)$$. The midpoint of $$BC$$ is $$M = \left(\frac{2+3}{2},\frac{2-2}{2},\frac{4+5}{2}\right) = \left(\frac{5}{2},0,\frac{9}{2}\right)$$.

Writing $$A = (1,4,3)$$, the length of the median $$AM$$ is $$AM = \sqrt{\left(\frac{5}{2}-1\right)^2 + (0-4)^2 + \left(\frac{9}{2}-3\right)^2} = \sqrt{\left(\frac{3}{2}\right)^2 + 16 + \left(\frac{3}{2}\right)^2} = \sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \sqrt{\frac{9 + 64 + 9}{4}} = \frac{\sqrt{82}}{2}\,. $$

Therefore, the answer is Option A: $$\frac{\sqrt{82}}{2}$$.