Let the vertex $$A$$ of a triangle $$ABC$$ be $$(1, 2)$$, and the mid-point of the side $$AB$$ be $$(5, -1)$$. If the centroid of this triangle is $$(3, 4)$$ and its circumcenter is $$(\alpha, \beta)$$, then $$21(\alpha + \beta)$$ is equal to :

The given data are:
A $$\left(1,2\right)$$, mid-point of $$AB$$ is $$\left(5,-1\right)$$, centroid $$G\left(3,4\right)$$.

1. Find the co-ordinates of $$B$$.
If $$B\left(x_B,y_B\right)$$, then using the mid-point formula
$$\frac{1+x_B}{2}=5,\;\; \frac{2+y_B}{2}=-1$$
$$\Rightarrow 1+x_B=10\;\; \text{and}\;\; 2+y_B=-2$$
$$\Rightarrow x_B=9,\;\; y_B=-4$$
Hence $$B\left(9,-4\right)$$.

2. Find the co-ordinates of $$C$$ from the centroid.
For $$C\left(x_C,y_C\right)$$, the centroid condition is
$$\frac{1+9+x_C}{3}=3,\;\; \frac{2-4+y_C}{3}=4$$
$$\Rightarrow 10+x_C=9\;\; \text{and}\;\; -2+y_C=12$$
$$\Rightarrow x_C=-1,\;\; y_C=14$$
Thus $$C\left(-1,14\right)$$.

3. Equation of the perpendicular bisector of $$AB$$.
Slope of $$AB$$: $$m_{AB}=\frac{-4-2}{9-1}=-\frac34$$
Perpendicular slope: $$\frac43$$.
Mid-point of $$AB$$ is $$\left(5,-1\right)$$, so
$$y+1=\frac43\bigl(x-5\bigr)\;\;-(1)$$

4. Equation of the perpendicular bisector of $$AC$$.
Slope of $$AC$$: $$m_{AC}=\frac{14-2}{-1-1}=-6$$
Perpendicular slope: $$\frac16$$.
Mid-point of $$AC$$ is $$\left(0,8\right)$$, so
$$y-8=\frac16\,x\;\;-(2)$$
or $$y=\frac16\,x+8$$.

5. Intersection of the two bisectors gives the circumcenter $$(\alpha,\beta)$$.
From $$(1)$$: $$y=\frac43\,(x-5)-1$$.
Set equal to $$(2)$$:
$$\frac43\,(x-5)-1=\frac16\,x+8$$
Multiply by $$6$$: $$8(x-5)-6=x+48$$
$$8x-40-6=x+48$$
$$8x-46=x+48$$
$$7x=94\;\;\Rightarrow\;\; x=\frac{94}{7}$$
Substitute in $$(2)$$:
$$y=\frac16\left(\frac{94}{7}\right)+8=\frac{94}{42}+8=\frac{47}{21}+8=\frac{215}{21}$$
Hence $$\alpha=\frac{94}{7},\;\; \beta=\frac{215}{21}$$.

6. Required value.
$$\alpha+\beta=\frac{94}{7}+\frac{215}{21}=\frac{282+215}{21}=\frac{497}{21}$$
Therefore $$21(\alpha+\beta)=21\cdot\frac{497}{21}=497$$.

Thus the correct choice is
Option C which is: $$497$$.