A ray of light passing through the point $$P(2, 3)$$ reflects on the $$X$$-axis at point $$A$$ and the reflected ray passes through the point $$Q(5, 4)$$. Let $$R$$ be the point that divides the line segment $$AQ$$ internally into the ratio $$2 : 1$$. Let the co-ordinates of the foot of the perpendicular $$M$$ from $$R$$ on the bisector of the angle $$PAQ$$ be $$(\alpha, \beta)$$. Then, the value of $$7\alpha + 3\beta$$ is equal to ______

A ray from $$P(2, 3)$$ reflects off the X-axis at point $$A$$ and passes through $$Q(5, 4)$$.

By the reflection principle, the reflected ray from $$A$$ to $$Q$$ corresponds to a direct ray from the reflection of $$P$$ in the X-axis, which is $$P'(2, -3)$$, to $$Q(5, 4)$$.

Line $$P'Q$$: slope $$= \frac{4-(-3)}{5-2} = \frac{7}{3}$$.

Equation: $$y + 3 = \frac{7}{3}(x - 2) \implies 3y + 9 = 7x - 14 \implies 7x - 3y = 23$$

Point $$A$$ is where this line meets the X-axis ($$y = 0$$):

$$7x = 23 \implies x = \frac{23}{7}$$

So $$A = \left(\frac{23}{7}, 0\right)$$.

Point $$R$$ divides $$AQ$$ internally in ratio $$2:1$$:

$$R = \left(\frac{2 \times 5 + 1 \times \frac{23}{7}}{3}, \frac{2 \times 4 + 1 \times 0}{3}\right) = \left(\frac{10 + \frac{23}{7}}{3}, \frac{8}{3}\right)$$

$$= \left(\frac{\frac{70+23}{7}}{3}, \frac{8}{3}\right) = \left(\frac{93}{21}, \frac{8}{3}\right) = \left(\frac{31}{7}, \frac{8}{3}\right)$$

The bisector of angle $$PAQ$$: Since $$A$$ is on the X-axis and the ray reflects off the X-axis, the angle bisector of $$\angle PAQ$$ is the Y-axis direction at $$A$$, i.e., the vertical line $$x = \frac{23}{7}$$.

The foot of perpendicular from $$R = \left(\frac{31}{7}, \frac{8}{3}\right)$$ to the line $$x = \frac{23}{7}$$ is:

$$M = \left(\frac{23}{7}, \frac{8}{3}\right)$$

So $$\alpha = \frac{23}{7}$$ and $$\beta = \frac{8}{3}$$.

$$7\alpha + 3\beta = 7 \times \frac{23}{7} + 3 \times \frac{8}{3} = 23 + 8 = 31$$

Hence the answer is $$\boxed{31}$$.