All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$2 \vec{OC} = 3 \vec{CA} = 3 (\vec{OA} - \vec{OC})$

$5 \vec{OC} = 3 a$

$\vec{OC} = \frac{3}{5} a$

$11 \vec{OD} = 5 \vec{OB}$

$\vec{OD} = \frac{5}{11} b$

(ii)
$\vec{BC} = \frac{3}{5}a -b$

$l_{BC}: r = b + \lambda (\frac{3}{5}a - b)$

$l_{BC}: r = \frac{3}{5} \lambda a + (1-\lambda)b, \lambda \in \mathbb{R}$

$\vec{AD} = \frac{3}{5}a -b$

$l_{AD}: r = a + \mu (\frac{5}{11}b - a)$

$l_{AD}: r = \frac{5}{11} \mu b + (1-\mu)a, \mu \in \mathbb{R}$

(iii)
At E, $\frac{5}{11} \mu b + (1-\mu)a = \frac{3}{5} \lambda a + (1-\lambda)b$

$\frac{5}{11} \mu = 1 - \lambda \rightarrow(1)$

$\frac{3}{5} \lambda = 1 - \mu \rightarrow(1)$

Solving with GC, $\lambda = \frac{3}{4}$ and $\mu = \frac{11}{20}$

$\vec{OC} = \frac{9}{20}a + \frac{1}{4} b$

$\vec{AE} = \frac{1}{4}b - \frac{11}{20}a$

$\vec{ED} = \frac{9}{44}b - \frac{9}{20}a$

$= \frac{9}{11}(\frac{1}{4}b - \frac{11}{20}a)$

$\vec{ED} = \frac{9}{11} \vec{AE}$

$\therefore, AE: ED = 11: 9$

(ii) should be manageable. For (iii), students need to be careful if they use a GC to solve the question. Its like statistics, solving $\mu$ and $\lambda$.