1. **Problem statement:** We have a parallelogram OACB with vectors \(\vec{a} = \overrightarrow{OA}\) and \(\vec{b} = \overrightarrow{OB}\). Points P and Q lie on OA and OB such that \(OP : PA = 2 : 1\) and \(BQ : QO = 3 : 1\). OPRQ is also a parallelogram. We need to: (i) Find the vector \(\overrightarrow{OR}\). (ii) Show that the area of parallelogram OACB is six times the area of parallelogram OPRQ. 2. **Find vectors for points P and Q:** - Since \(OP : PA = 2 : 1\), point P divides OA in ratio 2:1 from O to A. Thus, \(\overrightarrow{OP} = \frac{2}{3} \vec{a}\). - Since \(BQ : QO = 3 : 1\), point Q divides OB in ratio 3:1 from B to O. This means Q divides OB externally or internally? Since BQ:QO=3:1, Q lies on OB such that \(\overrightarrow{OQ} = \frac{1}{4} \vec{b}\) (because Q divides OB in ratio 1:3 from O to B). 3. **Find vector \(\overrightarrow{OR}\):** Since OPRQ is a parallelogram, \(\overrightarrow{OR} = \overrightarrow{OP} + \overrightarrow{OQ} = \frac{2}{3} \vec{a} + \frac{1}{4} \vec{b}\). 4. **Area comparison:** - Area of parallelogram OACB is \(|\vec{a} \times \vec{b}|\). - Area of parallelogram OPRQ is \(|\overrightarrow{OP} \times \overrightarrow{OQ}| = \left| \frac{2}{3} \vec{a} \times \frac{1}{4} \vec{b} \right| = \frac{2}{3} \times \frac{1}{4} |\vec{a} \times \vec{b}| = \frac{1}{6} |\vec{a} \times \vec{b}|\). Thus, area of OACB is 6 times area of OPRQ. **Final answers:** (i) \(\overrightarrow{OR} = \frac{2}{3} \vec{a} + \frac{1}{4} \vec{b}\) (ii) \(\text{Area}(OACB) = 6 \times \text{Area}(OPRQ)\)