1. **State the problem:** We have a ray of light passing through point $P(2, 3)$, reflecting on the x-axis at point $A$, and the reflected ray passes through point $Q(5, 4)$. Point $R$ divides the segment $AQ$ internally in ratio $2:1$. We need to find the coordinates $(\alpha, \beta)$ of the foot of the perpendicular $M$ from $R$ to the bisector of angle $PAQ$ and then compute $7\alpha + 3\beta$. 2. **Find point A:** Reflection about the x-axis changes the $y$ coordinate sign. Since $P$ maps to $Q$ after reflection at $A$, and $A$ lies on x-axis, coordinates of $A$ are $(x_A, 0)$. 3. **Use reflection property:** The incident ray from $P$ to $A$ and reflected ray from $A$ to $Q$ satisfy: Slope of $PA = \frac{0 - 3}{x_A - 2} = \frac{-3}{x_A - 2}$. Slope of $AQ = \frac{4 - 0}{5 - x_A} = \frac{4}{5 - x_A}$. For reflection on x-axis, the reflected slope is negative of incident slope: $$ \frac{4}{5 - x_A} = - \frac{-3}{x_A - 2} = \frac{3}{x_A - 2} $$ 4. **Solve for $x_A$: ** $$ \frac{4}{5 - x_A} = \frac{3}{x_A - 2} $$ Cross multiply: $$ 4 (x_A - 2) = 3 (5 - x_A) $$ $$ 4x_A - 8 = 15 - 3x_A $$ $$ 4x_A + 3x_A = 15 + 8 $$ $$ 7x_A = 23 $$ $$ x_A = \frac{23}{7} $$ So, $A = \left( \frac{23}{7}, 0 \right)$. 5. **Find point R dividing $A Q$ in ratio 2:1 internally:** Coordinates of $R$ are: $$ R = \left( \frac{2 \times 5 + 1 \times \frac{23}{7}}{2 + 1}, \frac{2 \times 4 + 1 \times 0}{3} \right) = \left( \frac{10 + \frac{23}{7}}{3}, \frac{8}{3} \right) = \left( \frac{\frac{70}{7} + \frac{23}{7}}{3}, \frac{8}{3} \right) = \left( \frac{93/7}{3}, \frac{8}{3} \right) = \left( \frac{93}{21}, \frac{8}{3} \right) = \left( \frac{31}{7}, \frac{8}{3} \right) $$ 6. **Find the bisector of angle $PAQ$: ** Vectors: $$ \vec{AP} = P - A = \left(2 - \frac{23}{7}, 3 - 0 \right) = \left( \frac{14 - 23}{7}, 3 \right) = \left( -\frac{9}{7}, 3 \right) $$ $$ \vec{AQ} = Q - A = \left(5 - \frac{23}{7}, 4 - 0 \right) = \left( \frac{35 - 23}{7}, 4 \right) = \left( \frac{12}{7}, 4 \right) $$ Normalize these vectors: $$ |\vec{AP}| = \sqrt{\left(-\frac{9}{7}\right)^2 + 3^2} = \sqrt{\frac{81}{49} + 9} = \sqrt{\frac{81 + 441}{49}} = \frac{\sqrt{522}}{7} = \frac{3\sqrt{58}}{7} $$ Unit vector $\hat{u}_{AP} = \left( -\frac{9}{7} \times \frac{7}{3\sqrt{58}}, 3 \times \frac{7}{3\sqrt{58}} \right) = \left( -\frac{9}{3\sqrt{58}}, \frac{7}{\sqrt{58}} \right) = \left( -\frac{3}{\sqrt{58}}, \frac{7}{\sqrt{58}} \right)$ Similarly for $\vec{AQ}$: $$ |\vec{AQ}| = \sqrt{ \left( \frac{12}{7} \right)^2 + 4^2 } = \sqrt{ \frac{144}{49} + 16 } = \sqrt{ \frac{144 + 784}{49} } = \frac{\sqrt{928}}{7} = \frac{4\sqrt{58}}{7} $$ Unit vector $\hat{u}_{AQ} = \left( \frac{12}{7} \times \frac{7}{4\sqrt{58}}, 4 \times \frac{7}{4\sqrt{58}} \right) = \left( \frac{12}{4\sqrt{58}}, \frac{7}{\sqrt{58}} \right) = \left( \frac{3}{\sqrt{58}}, \frac{7}{\sqrt{58}} \right)$ 7. **Vector bisector direction: ** Sum unit vectors: $$ \vec{d} = \hat{u}_{AP} + \hat{u}_{AQ} = \left( -\frac{3}{\sqrt{58}} + \frac{3}{\sqrt{58}}, \frac{7}{\sqrt{58}} + \frac{7}{\sqrt{58}} \right) = (0, \frac{14}{\sqrt{58}}) $$ So the angle bisector is vertical with direction vector $(0,1)$. 8. **Equation of angle bisector passing through $A$: ** Since $A = \left( \frac{23}{7}, 0 \right)$ and slope is infinite (vertical line), equation is: $$ x = \frac{23}{7} $$ 9. **Find foot of perpendicular $M = (\alpha, \beta)$ from $R \left( \frac{31}{7}, \frac{8}{3} \right)$ to line $x = \frac{23}{7}$: ** Since the line is vertical, the foot of the perpendicular has the same $x$ as the line and the same $y$ as $R$: $$ \alpha = \frac{23}{7}, \quad \beta = \frac{8}{3} $$ 10. **Final value:** $$ 7\alpha + 3\beta = 7 \times \frac{23}{7} + 3 \times \frac{8}{3} = 23 + 8 = 31 $$ **Answer: 31**