Let $\mathrm{G}$ be the centroid of a triangle $\mathrm{ABC}$ and $0 \mathrm{be}$ any other point in that…
Let $\mathrm{G}$ be the centroid of a triangle $\mathrm{ABC}$ and $0 \mathrm{be}$ any other point in that plane, then $\overline{\mathrm{OA}}+\overline{\mathrm{OB}}+\overline{\mathrm{OC}}+\overline{\mathrm{OG}}=$
$4 \overline{\mathrm{OG}}$
$\overline{\mathrm{O}}$
$3 \overline{\mathrm{OG}}$
$2 \overline{\mathrm{OG}}$
Solution
Let $\mathrm{O}$ be the origin.
Let $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of vertices $A, B, C$ respectively.
$\therefore \overline{\mathrm{OA}}+\overline{\mathrm{OB}}+\overline{\mathrm{OC}}=\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}$
Given : $G$ is the centroid of triangle
$\begin{array}{l}
\therefore \overline{\mathrm{OG}}=\frac{\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}}{3} \\
\therefore \overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}=3 \overline{\mathrm{OG}} \\
\therefore \overline{\mathrm{OA}}+\overline{\mathrm{OB}}+\overline{\mathrm{OC}}+\overline{\mathrm{OG}}=3 \overline{\mathrm{OG}}+\overline{\mathrm{OG}}=4 \overline{\mathrm{OG}}
\end{array}$