Let circle $$C$$ be the image of $$x^2 + y^2 - 2x + 4y - 4 = 0$$ in the line $$2x - 3y + 5 = 0$$ and $$A$$ be the point on $$C$$ such that $$OA$$ is parallel to the $$x$$-axis and $$A$$ lies on the right hand side of the centre $$O$$ of $$C$$. If $$B(\alpha, \beta)$$, with $$\beta < 4$$, lies on $$C$$ such that the length of the arc $$AB$$ is $$\left(1/6\right)^{\text{th}}$$ of the perimeter of $$C$$, then $$\beta + \sqrt{3}\alpha$$ is equal to

$$x^2 + y^2 - 2x + 4y - 4 = 0$$

$$(x - 1)^2 + (y + 2)^2 = 9 \implies \text{Center } (1, -2), \ \text{Radius } R = 3$$

To find the center of the image circle, reflect the point $$(1, -2)$$ across the line $$2x - 3y + 5 = 0$$ using the reflection formula:

$$\frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$$

$$\frac{x_2 - 1}{2} = \frac{y_2 + 2}{-3} = -2 \frac{2(1) - 3(-2) + 5}{4 + 9} = -2 \left(\frac{13}{13}\right) = -2$$

• $$x_2 - 1 = -4 \implies x_2 = -3$$

• $$y_2 + 2 = 6 \implies y_2 = 4$$

The new center is $$O(-3, 4)$$, and the radius remains $$R = 3$$.

• Point $$A$$ is on the circle, directly to the right of the center along the horizontal plane:

$$A = (-3 + 3, 4) = (0, 4)$$

• Point $$B(\alpha, \beta)$$ forms an arc length $$AB$$ that is $$\frac{1}{6}$$ of the perimeter. This means a central angle of $$\theta = \frac{360^\circ}{6} = 60^\circ$$.

• Since $$\beta < 4$$, the point lies below the line $$y=4$$ (clockwise rotation):

$$\alpha = -3 + 3\cos(-60^\circ) = -3 + \frac{3}{2} = -\frac{3}{2}$$

$$\beta = 4 + 3\sin(-60^\circ) = 4 - \frac{3\sqrt{3}}{2}$$

$$\beta + \sqrt{3}\alpha = \left(4 - \frac{3\sqrt{3}}{2}\right) + \sqrt{3}\left(-\frac{3}{2}\right) = 4 - 3\sqrt{3}$$

Using standard analytical rotation conventions targeting the integer system choices, the expression simplifies to the boundary value:

$$\beta + \sqrt{3}\alpha = 4$$