1. Find the point of reflection $A$ on the $x$-axis

Since the ray from $P(2,3)$ reflects on the $x$-axis and then passes through $Q(5,4)$, use the mirror-image method.

Reflect $Q(5,4)$ in the $x$-axis: $Q'(5,-4)$ Then $A$ is the intersection of the line $PQ'$ with the $x$-axis.

Line through $P(2,3)$ and $Q'(5,-4)$ has slope $m=\frac{-4-3}{5-2}= -\frac{7}{3}$ So its equation is $y-3=-\frac{7}{3}(x-2)$ At the $x$-axis, $y=0$: $-3=-\frac{7}{3}(x-2)$ $9=7(x-2)$ $7x=23$ $x=\frac{23}{7}$ Hence, $A\left(\frac{23}{7},0\right)$

2. Find the point $R$ dividing $AQ$ internally in the ratio $2:1$

$R$ divides $AQ$ internally in the ratio $2:1$, i.e. $AR:RQ=2:1$. Using section formula with $A\left(\frac{23}{7},0\right)$ and $Q(5,4)$, $R=\left(\frac{2\cdot 5+1\cdot \frac{23}{7}}{2+1},\frac{2\cdot 4+1\cdot 0}{3}\right)$ $R=\left(\frac{10+\frac{23}{7}}{3},\frac{8}{3}\right)$ $R=\left(\frac{\frac{93}{7}}{3},\frac{8}{3}\right)=\left(\frac{31}{7},\frac{8}{3}\right)$

3. Find the bisector of $\angle PAQ$

At point $A$, the lines are $AP$ and $AQ$.

Slope of $AP$: $m_{AP}=\frac{3-0}{2-\frac{23}{7}}=\frac{3}{-\frac{9}{7}}=-\frac{7}{3}$

Slope of $AQ$: $m_{AQ}=\frac{4-0}{5-\frac{23}{7}}=\frac{4}{\frac{12}{7}}=\frac{7}{3}$

Thus, the two rays through $A$ are symmetric about the vertical line through $A$. Therefore the bisector of $\angle PAQ$ is $x=\frac{23}{7}$

4. Find the foot of perpendicular from $R$ to this bisector

The bisector is the vertical line $x=\frac{23}{7}$ A perpendicular from $R\left(\frac{31}{7},\frac{8}{3}\right)$ to a vertical line is horizontal, so the foot $M(\alpha,\beta)$ has $\alpha=\frac{23}{7},\qquad \beta=\frac{8}{3}$

5. Compute $7\alpha+3\beta$

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

Therefore, $\boxed{31}$