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 :

The given circle is $$x^{2}+y^{2}+x-3y=0$$.
Complete the squares to obtain its centre and radius:

$$(x^{2}+x+\tfrac14)+(y^{2}-3y+\tfrac94)=\tfrac14+\tfrac94=\tfrac{5}{2}$$
$$\Rightarrow (x+\tfrac12)^{2}+(y-\tfrac32)^{2}=\tfrac{5}{2}$$
Centre $$C\!:\!\left(-\tfrac12,\tfrac32\right)$$, radius $$r=\sqrt{\tfrac{5}{2}}$$.

The point from which the chords are drawn is $$P(1,2)$$; verify that it lies on the circle:

$$1^{2}+2^{2}+1-3\cdot2=1+4+1-6=0$$, hence $$P$$ is on the circle.

Let a required chord have midpoint $$M(0,m)$$ on the $$y$$-axis. For the chord $$PS$$ with endpoints $$P(1,2)$$ and $$S(x_s,y_s)$$, the midpoint condition gives

$$\frac{1+x_s}{2}=0\;\Rightarrow\;x_s=-1,$$
$$\frac{2+y_s}{2}=m\;\Rightarrow\;y_s=2m-2.$$

The other end $$S(-1,\,2m-2)$$ must satisfy the circle equation:

$$(-1)^{2}+(2m-2)^{2}+(-1)-3(2m-2)=0.$$

Simplify:

$$1+4(m-1)^{2}-1-6m+6=0$$
$$\Rightarrow 4(m^{2}-2m+1)-6m+6=0$$
$$\Rightarrow 4m^{2}-8m+4-6m+6=0$$
$$\Rightarrow 4m^{2}-14m+10=0$$
$$\Rightarrow 2m^{2}-7m+5=0.$$

Solve the quadratic:

$$\Delta = (-7)^{2}-4\cdot2\cdot5 = 49-40 = 9,$$
$$m=\frac{7\pm3}{4}\;\Longrightarrow\;m_1=\frac{10}{4}=\frac52,\;m_2=\frac{4}{4}=1.$$

Thus there are two such chords:

Case 1: $$m=\tfrac52 \;\Rightarrow\; S_1(-1,3).$$
Case 2: $$m=1 \;\Rightarrow\; S_2(-1,0).$$

Denote the two other ends by $$R(-1,3)$$ and $$S(-1,0)$$. The midpoint $$\left(\alpha,\beta\right)$$ of $$RS$$ is

$$\alpha=\frac{-1+(-1)}{2}=-1,\qquad \beta=\frac{3+0}{2}=\frac32.$$

Hence

$$6(\alpha+\beta)=6\left(-1+\frac32\right)=6\left(\frac12\right)=3.$$

Therefore, $$6(\alpha+\beta)=3$$.

Option B which is: $$3$$