Let the vertex A of a triangle ABC be (1, 2) , and the mid-point of...

Let the coordinates of vertex $B$ be $(x_1, y_1) and vertex$C$be$(x_2, y_2)$.<br><br>Since the mid-point of$AB$is$(5, -1)$, we have:<br>$\dfrac{1 + x_1}{2} = 5 \Rightarrow x_1 = 9$<br>$\dfrac{2 + y_1}{2} = -1 \Rightarrow y_1 = -4$<br>Thus,$B \equiv (9, -4)$.<br><br>Given the centroid of$\triangle ABC$is$(3, 4)$, we get:<br>$\dfrac{1 + 9 + x_2}{3} = 3 \Rightarrow x_2 = -1$<br>$\dfrac{2 - 4 + y_2}{3} = 4 \Rightarrow y_2 = 14$<br>Thus,$C \equiv (-1, 14)$.<br><br>The circumcenter$(\alpha, \beta)$is the intersection of the perpendicular bisectors of the sides of the triangle.<br><br>For side$AB$, the mid-point is$(5, -1)$and the slope is$\dfrac{-4 - 2}{9 - 1} = -\dfrac{3}{4}$.<br>The slope of its perpendicular bisector is$\dfrac{4}{3}$.<br>Equation of the perpendicular bisector of$AB$is:<br>$y - (-1) = \dfrac{4}{3}(x - 5) \Rightarrow 4x - 3y = 23$<br><br>For side$AC$, the mid-point is \left(\dfrac{1 - 1}{2}, \dfrac{2 + 14}{2}\right) = (0, 8)$ and the slope is $\dfrac{14 - 2}{-1 - 1} = -6$.
The slope of its perpendicular bisector is $\dfrac{1}{6}$.
Equation of the perpendicular bisector of $AC$ is:
$y - 8 = \dfrac{1}{6}(x - 0) \Rightarrow x - 6y = -48$

Solving the equations $4x - 3y = 23$ and $x - 6y = -48$:
Multiplying the first equation by $2$ gives $8x - 6y = 46$.
Subtracting the second equation from this gives:
$7x = 94 \Rightarrow x = \dfrac{94}{7}$
Substituting $x$ into the second equation:
$\dfrac{94}{7} - 6y = -48 \Rightarrow 6y = \dfrac{430}{7} \Rightarrow y = \dfrac{215}{21}$

Therefore, the circumcenter is $(\alpha, \beta) = \left(\dfrac{94}{7}, \dfrac{215}{21}\right)$.

Finally, calculating $21(\alpha + \beta)$:
$21(\alpha + \beta) = 21\left(\dfrac{94}{7} + \dfrac{215}{21}\right) = 3(94) + 215 = 282 + 215 = 497$.

Answer: $497$