1. **State the problem:** Find the point $(a,b)$ on the line $2x - 3y + 6 = 0$ that is closest to the point $(3,1)$, then find $a+b$. 2. **Formula and concept:** The closest point on a line to a given point is the foot of the perpendicular from the point to the line. 3. **Line equation:** $2x - 3y + 6 = 0$ can be rewritten as $2x - 3y = -6$. 4. **Given point:** $P = (3,1)$. 5. **Find the perpendicular line:** The slope of the given line is $m = \frac{2}{3}$ (since $2x - 3y + 6=0 \Rightarrow y = \frac{2}{3}x + 2$). 6. The slope of the perpendicular line is the negative reciprocal: $m_\perp = -\frac{3}{2}$. 7. Equation of the perpendicular line through $(3,1)$: $$y - 1 = -\frac{3}{2}(x - 3)$$ which simplifies to $$y = -\frac{3}{2}x + \frac{9}{2} + 1 = -\frac{3}{2}x + \frac{11}{2}$$ 8. **Find intersection point $(a,b)$ of the two lines:** Solve the system: $$\begin{cases} 2x - 3y + 6 = 0 \\ y = -\frac{3}{2}x + \frac{11}{2} \end{cases}$$ Substitute $y$: $$2x - 3\left(-\frac{3}{2}x + \frac{11}{2}\right) + 6 = 0$$ $$2x + \frac{9}{2}x - \frac{33}{2} + 6 = 0$$ Multiply all terms by 2 to clear denominators: $$4x + 9x - 33 + 12 = 0$$ $$13x - 21 = 0$$ $$x = \frac{21}{13}$$ 9. Find $y$: $$y = -\frac{3}{2} \times \frac{21}{13} + \frac{11}{2} = -\frac{63}{26} + \frac{143}{26} = \frac{80}{26} = \frac{40}{13}$$ 10. So the closest point is: $$\left(\frac{21}{13}, \frac{40}{13}\right)$$ 11. Find $a+b$: $$a+b = \frac{21}{13} + \frac{40}{13} = \frac{61}{13}$$ **Final answer:** $\boxed{\frac{61}{13}}$