If $\overline{\mathrm{OA}}=\overline{\mathrm{a}}\text{ and }\overline{\text{O}\mathrm{B}}=\overline{\overline{\mathrm{b}}}$ then the position vectors of C and D where C is on AB produced such that $\overline{\mathrm{AC}}=3\overline{\mathrm{AB}}$ and D is on BA produced such that $\overline{\mathrm{BD}}=2\overline{\mathrm{BA}}$are respectively.

Given $\overline{\mathrm{OA}}=\overline{\mathrm{a}},\overline{\mathrm{OB}}=\overline{\mathrm{b}}$ where O is the origin of vectors, also given
$\begin{array}{l}\overline{\mathrm{AC}}=3\overline{\mathrm{AB}}\Rightarrow \overline{\mathrm{OC}}-\overline{\mathrm{OA}}=3(\overline{\mathrm{OB}}-\overline{\mathrm{OA}})\\ \Rightarrow \overline{\mathrm{OC}}=3\overline{\mathrm{OB}}-2\cdot \overline{\mathrm{OA}}=3\overline{\mathrm{b}}-2\overline{\mathrm{a}}\end{array}$
Again $\overline{\mathrm{BD}}=2\overline{\mathrm{BA}}\Rightarrow \overline{\mathrm{OD}}-\overline{\mathrm{OB}}=2(\overline{\mathrm{OA}}-\overline{\mathrm{OB}})\Rightarrow \overline{\mathrm{OD}}=2.\overline{\mathrm{OA}}-\overline{\mathrm{OB}}=2\overline{\mathrm{a}}-\overline{\mathrm{b}}$