Let A 3 , 0 , - 1 ,   B 2 , 10 , 6 and C 1 , 2 , 1 be the vertices of a triangle and M be the mid-point…

$\because$ $M$ is mid point of $AC \& G$ divides $BM$ in the ratio $2:1$ internally

$\therefore G$ is centroid of $△ABC$

$\therefore G=\left(\frac{3+1+2}{3},\frac{0+2+10}{3},\frac{-1+1+6}{3}\right)=\left(2,4,2\right)$

$\therefore \overrightarrow{OA}=3\overrightarrow{i}+0\overrightarrow{j}-\overrightarrow{k} \& \overrightarrow{OG}=2\overrightarrow{i}+4\overrightarrow{j}+2\overrightarrow{k}$

$\therefore \cos \angle GOA=\frac{\overrightarrow{OA}.\overrightarrow{OG}}{\left|\overrightarrow{OA}\right|\left|\overrightarrow{OG}\right|}=\frac{6-2}{\sqrt{10}\sqrt{24}}=\frac{4}{\sqrt{10} 2\sqrt{6}}$

$=\frac{1}{\sqrt{15}}$