1. **Problem statement:** We have quadrilateral OABC with vectors $\vec{OA} = \vec{a}$ and $\vec{OB} = \vec{b}$. Points M and N lie on lines OB and AB respectively, with ratios $OM : MB = 1 : 5$ and $AN : NB = 3 : 2$. Lines MNC and ANB are straight, and $\vec{OB}$ is parallel to $\vec{AC}$. We need to find the ratio $OB : AC$ in the form $k : 1$. 2. **Express points M and N in vector form:** - Since $OM : MB = 1 : 5$, point M divides OB in ratio 1:5, so $$\vec{OM} = \frac{1}{1+5} \vec{OB} = \frac{1}{6} \vec{b}.$$ - Since $AN : NB = 3 : 2$, point N divides AB in ratio 3:2, so $$\vec{AN} = \frac{3}{3+2} \vec{AB} = \frac{3}{5} (\vec{b} - \vec{a})$$ which gives $$\vec{ON} = \vec{OA} + \vec{AN} = \vec{a} + \frac{3}{5}(\vec{b} - \vec{a}) = \vec{a} + \frac{3}{5}\vec{b} - \frac{3}{5}\vec{a} = \frac{2}{5}\vec{a} + \frac{3}{5}\vec{b}.$$ 3. **Use the fact that M, N, C are collinear:** Since M, N, C lie on a straight line, vector $\vec{NC}$ is parallel to vector $\vec{NM}$. Express $\vec{NC} = \vec{OC} - \vec{ON}$ and $\vec{NM} = \vec{OM} - \vec{ON}$. Let $\vec{OC} = \vec{c}$. Then $$\vec{NC} = \vec{c} - \left( \frac{2}{5}\vec{a} + \frac{3}{5}\vec{b} \right),$$ $$\vec{NM} = \frac{1}{6}\vec{b} - \left( \frac{2}{5}\vec{a} + \frac{3}{5}\vec{b} \right) = -\frac{2}{5}\vec{a} + \left( \frac{1}{6} - \frac{3}{5} \right) \vec{b} = -\frac{2}{5}\vec{a} - \frac{13}{30}\vec{b}.$$ Since $\vec{NC}$ is parallel to $\vec{NM}$, there exists a scalar $\lambda$ such that $$\vec{c} - \frac{2}{5}\vec{a} - \frac{3}{5}\vec{b} = \lambda \left(-\frac{2}{5}\vec{a} - \frac{13}{30}\vec{b} \right).$$ Rearranged, $$\vec{c} = \frac{2}{5}\vec{a} + \frac{3}{5}\vec{b} + \lambda \left(-\frac{2}{5}\vec{a} - \frac{13}{30}\vec{b} \right) = \left( \frac{2}{5} - \frac{2}{5} \lambda \right) \vec{a} + \left( \frac{3}{5} - \frac{13}{30} \lambda \right) \vec{b}.$$ 4. **Use the fact that $\vec{OB}$ is parallel to $\vec{AC}$:** Vector $\vec{AC} = \vec{OC} - \vec{OA} = \vec{c} - \vec{a}$. Since $\vec{OB} = \vec{b}$ is parallel to $\vec{AC}$, there exists a scalar $k$ such that $$\vec{c} - \vec{a} = k \vec{b}.$$ Substitute $\vec{c}$ from above: $$\left( \frac{2}{5} - \frac{2}{5} \lambda \right) \vec{a} + \left( \frac{3}{5} - \frac{13}{30} \lambda \right) \vec{b} - \vec{a} = k \vec{b}.$$ Simplify the $\vec{a}$ terms: $$\left( \frac{2}{5} - \frac{2}{5} \lambda - 1 \right) \vec{a} + \left( \frac{3}{5} - \frac{13}{30} \lambda \right) \vec{b} = k \vec{b}.$$ Since $\vec{a}$ and $\vec{b}$ are independent, the coefficient of $\vec{a}$ must be zero: $$\frac{2}{5} - \frac{2}{5} \lambda - 1 = 0 \implies \frac{2}{5} - 1 = \frac{2}{5} \lambda \implies -\frac{3}{5} = \frac{2}{5} \lambda \implies \lambda = -\frac{3}{5} \times \frac{5}{2} = -\frac{3}{2}.$$ 5. **Find $k$ using $\lambda = -\frac{3}{2}$:** $$k = \frac{3}{5} - \frac{13}{30} \times \left(-\frac{3}{2} \right) = \frac{3}{5} + \frac{39}{60} = \frac{3}{5} + \frac{13}{20} = \frac{12}{20} + \frac{13}{20} = \frac{25}{20} = \frac{5}{4}.$$ 6. **Find the ratio $OB : AC$:** Since $\vec{AC} = k \vec{b}$ and $\vec{OB} = \vec{b}$, the magnitudes satisfy $$OB : AC = 1 : k = 1 : \frac{5}{4} = \frac{4}{5} : 1.$$ **Final answer:** $$\boxed{OB : AC = \frac{4}{5} : 1}.$$