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:

Let the coordinates of vertex $B$ be $(x_1, y_1)$ and vertex $C$ be $(x_2, y_2)$.

Since the mid-point of $AB$ is $(5, -1)$, we have:
$\dfrac{1 + x_1}{2} = 5 \Rightarrow x_1 = 9$
$\dfrac{2 + y_1}{2} = -1 \Rightarrow y_1 = -4$
Thus, $B \equiv (9, -4)$.

Given the centroid of $\triangle ABC$ is $(3, 4)$, we get:
$\dfrac{1 + 9 + x_2}{3} = 3 \Rightarrow x_2 = -1$
$\dfrac{2 - 4 + y_2}{3} = 4 \Rightarrow y_2 = 14$
Thus, $C \equiv (-1, 14)$.

The circumcenter $(\alpha, \beta)$ is the intersection of the perpendicular bisectors of the sides of the triangle.

For side $AB$, the mid-point is $(5, -1)$ and the slope is $\dfrac{-4 - 2}{9 - 1} = -\dfrac{3}{4}$.
The slope of its perpendicular bisector is $\dfrac{4}{3}$.
Equation of the perpendicular bisector of $AB$ is:
$y - (-1) = \dfrac{4}{3}(x - 5) \Rightarrow 4x - 3y = 23$

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: $497$