Suppose that two chords, drawn from the point $(1, 2)$ on the circle $x^2 + y^2 + x - 3y = 0$ are bisected by the $y$-axis. If the other ends of these chords are $R$ and $S$, and the mid point of the line segment $RS$ is $(\alpha, \beta)$, then $6(\alpha + \beta)$ is equal to:

Let the other end of the chord be $(x_1, y_1)$.

Since the chord is bisected by the $y$-axis, the midpoint of the chord lies on the $y$-axis.

The midpoint of the chord joining $(1, 2)$ and $(x_1, y_1)$ is $\left(\dfrac{x_1 + 1}{2}, \dfrac{y_1 + 2}{2}\right)$.

Since it lies on the $y$-axis, its $x$-coordinate must be zero:

$\dfrac{x_1 + 1}{2} = 0 \Rightarrow x_1 = -1$

Since the point $(x_1, y_1)$ lies on the given circle $x^2 + y^2 + x - 3y = 0$, we substitute $x_1 = -1$ into the equation:

$(-1)^2 + y_1^2 + (-1) - 3y_1 = 0$

$1 + y_1^2 - 1 - 3y_1 = 0$

$y_1^2 - 3y_1 = 0$

$y_1(y_1 - 3) = 0 \Rightarrow y_1 = 0$ or $y_1 = 3$

Thus, the other ends of the two chords are $R(-1, 0)$ and $S(-1, 3)$.

The midpoint of the line segment $RS$ is given as $(\alpha, \beta)$.

$\alpha = \dfrac{-1 + (-1)}{2} = -1$

$\beta = \dfrac{0 + 3}{2} = \dfrac{3}{2}$

We need to find the value of $6(\alpha + \beta)$:

$6(\alpha + \beta) = 6\left(-1 + \dfrac{3}{2}\right) = 6\left(\dfrac{1}{2}\right) = 3$

Answer: $3$