Suppose that two chords, drawn from the point (1, 2) on the circle...

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)$.<br><br>Since it lies on the$y$-axis, its$x$-coordinate must be zero:<br><br>$\dfrac{x_1 + 1}{2} = 0 \Rightarrow x_1 = -1$<br><br>Since the point$(x_1, y_1)$lies on the given circle$x^2 + y^2 + x - 3y = 0$, we substitute$x_1 = -1$into the equation:<br><br>$(-1)^2 + y_1^2 + (-1) - 3y_1 = 0$<br><br>$1 + y_1^2 - 1 - 3y_1 = 0$<br><br>$y_1^2 - 3y_1 = 0$<br><br>$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: $3$