1. **Stating the problem:** We have a quadrilateral OABC with vectors \(\overrightarrow{OA} = \mathbf{a}\), \(\overrightarrow{OB} = \mathbf{b}\), and \(\overrightarrow{OC} = \mathbf{c}\). Point \(P\) lies on line AC such that \(AP : PC = 3 : 2\), and OB and AC intersect at \(P\). We are to find \(\overrightarrow{OP}\) in terms of \(\mathbf{a}\) and \(\mathbf{c}\). 2. **Express the position vector of P on AC:** Since \(P\) divides \(AC\) in the ratio \(3:2\), using section formula, $$\overrightarrow{OP} = \frac{3\overrightarrow{C} + 2\overrightarrow{A}}{3+2} = \frac{3\mathbf{c} + 2\mathbf{a}}{5}.$$ Here, \(\overrightarrow{A} = \mathbf{a}\), \(\overrightarrow{C} = \mathbf{c}\). 3. **Therefore, the position vector of point P is:** $$\boxed{\overrightarrow{OP} = \frac{2}{5}\mathbf{a} + \frac{3}{5}\mathbf{c}}.$$ **Explanation:** We use the section formula for internal division of a line segment in ratio 3:2. The vector \(\overrightarrow{OP}\) is a weighted average of vectors \(\mathbf{a}\) and \(\mathbf{c}\), weighted by the ratio parts.