1. We are given points A(11,-2), B(3,6), and a point P(a,3) on the line joining A and B. 2. We want to find the ratio in which P divides AB and also the value of a. 3. Let the ratio be $k:1$ such that P divides AB internally. 4. The formula for the coordinates of P dividing AB in ratio $k:1$ is $$\left(\frac{11k+3}{k+1}, \frac{-2k+6}{k+1}\right).$$ 5. We know the y-coordinate of P is 3, so $$\frac{-2k+6}{k+1} = 3.$$ Multiply both sides by $(k+1)$: $$-2k + 6 = 3k + 3.$$ 6. Rearranging, $$6 - 3 = 3k + 2k,$$ so $$3 = 5k.$$ Therefore, $$k = \frac{3}{5}.$$ 7. Using $k=\frac{3}{5}$ to find the x-coordinate $a$ of P: $$a = \frac{11k + 3}{k+1} = \frac{11 \times \frac{3}{5} + 3}{\frac{3}{5} + 1} = \frac{\frac{33}{5} + 3}{\frac{8}{5}} = \frac{\frac{33}{5} + \frac{15}{5}}{\frac{8}{5}} = \frac{\frac{48}{5}}{\frac{8}{5}} = \frac{48}{5} \times \frac{5}{8} = 6.$$ 8. Hence, the point P divides AB in ratio $3:5$ and $a=6$.