1. **Problem statement:** Given quadrilateral OABC with vectors \(\vec{OA} = \vec{a}\), \(\vec{AB} = 3\vec{b} + \frac{1}{2}\vec{a}\), and \(\vec{OC} = 2\vec{b}\). Point D lies on OB such that \(\vec{OD} = \frac{3}{5} \vec{OB}\). Point E lies on CB such that \(\vec{CE} = k \vec{CB}\). Given that points A, D, and E are collinear, find (i) the value of \(k\) and (ii) the ratio \(AD : DE\). 2. **Find vectors \(\vec{OB}\) and \(\vec{CB}\):** \[ \vec{OB} = \vec{OA} + \vec{AB} = \vec{a} + \left(3\vec{b} + \frac{1}{2}\vec{a}\right) = \frac{3}{2}\vec{a} + 3\vec{b} \] \[ \vec{CB} = \vec{OB} - \vec{OC} = \left(\frac{3}{2}\vec{a} + 3\vec{b}\right) - 2\vec{b} = \frac{3}{2}\vec{a} + \vec{b} \] 3. **Express \(\vec{OD}\) and \(\vec{AD}\):** \[ \vec{OD} = \frac{3}{5} \vec{OB} = \frac{3}{5} \left(\frac{3}{2}\vec{a} + 3\vec{b}\right) = \frac{9}{10}\vec{a} + \frac{9}{5}\vec{b} \] \[ \vec{AD} = \vec{OD} - \vec{OA} = \left(\frac{9}{10}\vec{a} + \frac{9}{5}\vec{b}\right) - \vec{a} = \left(\frac{9}{10} - 1\right)\vec{a} + \frac{9}{5}\vec{b} = -\frac{1}{10}\vec{a} + \frac{9}{5}\vec{b} \] 4. **Express \(\vec{AE}\) in terms of \(k\):** \[ \vec{AE} = \vec{OE} - \vec{OA} \] Since \(\vec{OE} = \vec{OC} + \vec{CE} = 2\vec{b} + k \left(\frac{3}{2}\vec{a} + \vec{b}\right) = \frac{3k}{2}\vec{a} + (2 + k)\vec{b} \), \[ \vec{AE} = \left(\frac{3k}{2}\vec{a} + (2 + k)\vec{b}\right) - \vec{a} = \left(\frac{3k}{2} - 1\right)\vec{a} + (2 + k)\vec{b} \] 5. **Since A, D, E are collinear, \(\vec{AE} = \lambda \vec{AD}\) for some scalar \(\lambda\):** Equate components: \[ \left(\frac{3k}{2} - 1\right)\vec{a} + (2 + k)\vec{b} = \lambda \left(-\frac{1}{10}\vec{a} + \frac{9}{5}\vec{b}\right) \] This gives two equations: \[ \frac{3k}{2} - 1 = -\frac{\lambda}{10} \quad (1) \] \[ 2 + k = \frac{9\lambda}{5} \quad (2) \] 6. **Solve for \(k\) and \(\lambda\):** From (1): \[ \lambda = -10 \left(\frac{3k}{2} - 1\right) = -15k + 10 \] Substitute into (2): \[ 2 + k = \frac{9}{5}(-15k + 10) = -27k + 18 \] \[ 2 + k + 27k = 18 \Rightarrow 28k = 16 \Rightarrow k = \frac{16}{28} = \frac{4}{7} \] 7. **Find \(\lambda\):** \[ \lambda = -15 \times \frac{4}{7} + 10 = -\frac{60}{7} + 10 = \frac{10 \times 7 - 60}{7} = \frac{70 - 60}{7} = \frac{10}{7} \] 8. **Find ratio \(AD : DE\):** Since \(\vec{AE} = \vec{AD} + \vec{DE} = \lambda \vec{AD}\), \[ \vec{DE} = \vec{AE} - \vec{AD} = (\lambda - 1) \vec{AD} \] So, \[ AD : DE = 1 : (\lambda - 1) = 1 : \left(\frac{10}{7} - 1\right) = 1 : \frac{3}{7} = 7 : 3 \] **Final answers:** (i) \(k = \frac{4}{7}\) (ii) \(AD : DE = 7 : 3\)