Consider a triangle having vertices $$A(-2, 3)$$, $$B(1, 9)$$ and $$C(3, 8)$$. If a line $$L$$ passing through the circumcentre of triangle $$ABC$$, bisects line $$BC$$, and intersects y-axis at point $$\left(0, \frac{\alpha}{2}\right)$$, then the value of real number $$\alpha$$ is ___.

The vertices of the triangle are $$A(-2, 3)$$, $$B(1, 9)$$, and $$C(3, 8)$$. We first find the circumcentre of the triangle.

The circumcentre $$O = (h, k)$$ is equidistant from all three vertices. From $$|OA|^2 = |OB|^2$$: $$(h+2)^2 + (k-3)^2 = (h-1)^2 + (k-9)^2$$ $$h^2 + 4h + 4 + k^2 - 6k + 9 = h^2 - 2h + 1 + k^2 - 18k + 81$$ $$6h + 12k = 69 \implies 2h + 4k = 23 \quad \cdots (1)$$

From $$|OB|^2 = |OC|^2$$: $$(h-1)^2 + (k-9)^2 = (h-3)^2 + (k-8)^2$$ $$h^2 - 2h + 1 + k^2 - 18k + 81 = h^2 - 6h + 9 + k^2 - 16k + 64$$ $$4h - 2k = -9 \quad \cdots (2)$$

From (1): $$h = \frac{23 - 4k}{2}$$. Substituting into (2): $$4 \cdot \frac{23 - 4k}{2} - 2k = -9$$ $$2(23 - 4k) - 2k = -9$$ $$46 - 8k - 2k = -9$$ $$10k = 55 \implies k = \frac{11}{2}$$

Then $$h = \frac{23 - 22}{2} = \frac{1}{2}$$. So the circumcentre is $$O = \left(\frac{1}{2}, \frac{11}{2}\right)$$.

The midpoint of $$BC$$ is $$M = \left(\frac{1+3}{2}, \frac{9+8}{2}\right) = \left(2, \frac{17}{2}\right)$$.

The line $$L$$ passes through $$O = \left(\frac{1}{2}, \frac{11}{2}\right)$$ and $$M = \left(2, \frac{17}{2}\right)$$. The slope is: $$m = \frac{\frac{17}{2} - \frac{11}{2}}{2 - \frac{1}{2}} = \frac{3}{\frac{3}{2}} = 2$$

The equation of line $$L$$: $$y - \frac{11}{2} = 2\left(x - \frac{1}{2}\right) \Rightarrow y = 2x + \frac{11}{2} - 1 = 2x + \frac{9}{2}$$.

At $$x = 0$$: $$y = \frac{9}{2}$$. The y-intercept is $$\left(0, \frac{9}{2}\right) = \left(0, \frac{\alpha}{2}\right)$$, so $$\alpha = 9$$.