Let the line $$x - y = 4$$ intersect the circle $$C: (x - 4)^2 + (y + 3)^2 = 9$$ at the points Q and R. If $$P(\alpha, \beta)$$ is a point on C such that $$PQ = PR$$, then $$(6\alpha + 8\beta)^2$$ is equal to __________.

If a point $$P$$ on a circle is equidistant from two points $$Q$$ and $$R$$ on the same circle ($$PQ = PR$$), then $$P$$ must lie on the perpendicular bisector of the chord $$QR$$. 

The perpendicular bisector of any chord of a circle always passes through the center of the circle.Therefore, the line $$OP$$ (where $$O$$ is the center) is perpendicular to the line $$QR$$.

Center $$O = (4, -3)$$ and Radius $$r = 3$$

The given chord line is $$x - y = 4 \implies y = x - 4$$. Slope of the chord $$QR$$ is $$m_1 = 1$$.

Slope of $$OP$$ is $$m_2 = -\frac{1}{m_1} = -1$$

$$y - (-3) = -1(x - 4)$$ 

$$y + 3 = -x + 4 \implies x + y = 1$$

$$\alpha + \beta = 1 \implies \beta = 1 - \alpha$$

$$(\alpha - 4)^2 + (\beta + 3)^2 = 9$$

$$(\alpha - 4)^2 + (1 - \alpha + 3)^2 = 9$$

$$2(\alpha - 4)^2 = 9 \implies (\alpha - 4)^2 = \frac{9}{2}$$

$$\alpha - 4 = \pm \frac{3}{\sqrt{2}} \implies \alpha = 4 \pm \frac{3}{\sqrt{2}}$$

$$\beta = 1 - \alpha = 1 - \left(4 \pm \frac{3}{\sqrt{2}}\right) = -3 \mp \frac{3}{\sqrt{2}}$$

$$6\alpha + 8\beta = 6\left(4 \pm \frac{3}{\sqrt{2}}\right) + 8\left(-3 \mp \frac{3}{\sqrt{2}}\right)$$

$$6\alpha + 8\beta = 24 \pm \frac{18}{\sqrt{2}} - 24 \mp \frac{24}{\sqrt{2}}$$

$$6\alpha + 8\beta = \mp \frac{6}{\sqrt{2}} = \mp 3\sqrt{2}$$

$$(6\alpha + 8\beta)^2 = (\mp 3\sqrt{2})^2 = 9 \times 2 = 18$$