If \(D, E\) and \(F\) are the mid-points of the sides \(B C\), \(\mathrm{CA}\) and \(\mathrm{AB}\) of triangle \(\mathrm{ABC}\) respectively, then \(\overline{\mathrm{AD}}+\frac{2}{3} \overline{\mathrm{BE}}+\frac{1}{3} \overline{\mathrm{CF}}=\)

Let the position vector of A, B, C, D, E, F be \(\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}, \overline{\mathrm{d}}, \overline{\mathrm{e}}, \overline{\mathrm{f}}\) respectively.
\(\therefore \quad \overline{\mathrm{d}} =\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}, \overline{\mathrm{e}}=\frac{\overline{\mathrm{c}}+\overline{\mathrm{a}}}{2}, \overline{\mathrm{f}}=\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2} \)
\( \text { Now, } \overline{\mathrm{AD}}+\frac{2}{3} \overline{\mathrm{BE}}+\frac{1}{3} \overline{\mathrm{CF}} \)
\( =\overline{\mathrm{d}}-\overline{\mathrm{a}}+\frac{2}{3}(\overline{\mathrm{e}}-\overline{\mathrm{b}})+\frac{1}{3}(\overline{\mathrm{f}}-\overline{\mathrm{c}}) \)
\( =\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}}{2}-\overline{\mathrm{a}}+\frac{2}{3}\left(\frac{\overline{\mathrm{c}}+\overline{\mathrm{a}}}{2}-\overline{\mathrm{b}}\right)\) \(+\frac{1}{3}\left(\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}}{2}-\overline{\mathrm{c}}\right) \)
\( =\frac{\overline{\mathrm{b}}+\overline{\mathrm{c}}-2 \overline{\mathrm{a}}}{2}+\frac{\overline{\mathrm{c}}+\overline{\mathrm{a}}-2 \overline{\mathrm{b}}}{3}+\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}-2 \overline{\mathrm{c}}}{6} \)
\( =\frac{3 \overline{\mathrm{c}}-3 \overline{\mathrm{a}}}{6} \)
\( =\frac{3}{6}(\overline{\mathrm{c}}-\overline{\mathrm{a}}) \)
\( =\frac{1}{2} \overline{\mathrm{AC}} \)