1. Write the coordinates of the points

From the given position vectors,

$A=(1,4,3),\quad B=(2,\alpha,4),\quad C=(3,-2,5).$

2. Use the condition for collinearity

The points $A,B,C$ are collinear if vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel.

Compute:

$\overrightarrow{AB}=B-A=(2-1,\alpha-4,4-3)=(1,\alpha-4,1),$

$\overrightarrow{AC}=C-A=(3-1,-2-4,5-3)=(2,-6,2).$

For collinearity, we need

$\overrightarrow{AB}=\lambda \overrightarrow{AC}.$

Comparing coordinates:

$1=2\lambda \implies \lambda=\frac12.$

Then from the second coordinate,

$\alpha-4=-6\left(\frac12\right)=-3$

so

$\alpha=1.$

Thus, when $\alpha=1$, the points are collinear.

We need the smallest positive integer for which they are non-collinear. Since $\alpha=1$ makes them collinear, the smallest positive integer making them non-collinear is

$\alpha=2.$

3. Substitute $\alpha=2$

Then

$B=(2,2,4).$

4. Find the midpoint of $BC$

Let $M$ be the midpoint of $BC$. Then

$M=\left(\frac{2+3}{2},\frac{2+(-2)}{2},\frac{4+5}{2}\right)=\left(\frac52,0,\frac92\right).$

5. Find the median through $A$

The median through $A$ is $AM$.

$\overrightarrow{AM}=M-A=\left(\frac52-1,0-4,\frac92-3\right)=\left(\frac32,-4,\frac32\right).$

Its length is

$AM=\sqrt{\left(\frac32\right)^2+(-4)^2+\left(\frac32\right)^2}$

$=\sqrt{\frac94+16+\frac94}$

$=\sqrt{\frac{9+64+9}{4}}=\sqrt{\frac{82}{4}}=\frac{\sqrt{82}}{2}.$

6. Match with the options

$\boxed{\frac{\sqrt{82}}{2}}$

So the correct option is A.