If $G$ be the centroid of the $∆ABC$ and $O$ be any other point in the plane of the triangle $ABC$, then 

Let $G$ be the origin and $GO$ be $X$-axis and $O\equiv (a,0),A\equiv \left(x_1,y_1\right),B\equiv \left(x_2,y_2\right)$$,C\equiv \left(x_3,y_3\right)$

$\text{ Now, LHS }=O{A}^2+O{B}^2+O{C}^2$$={\left(x_1-a\right)}^2+y_1^2+{\left(x_2-a\right)}^2+y_2^2+{\left(x_3-a\right)}^2+y_3^2$

 $\Rightarrow LHS=\left(x_1^2+x_2^2+x_3^2\right)+\left(y_1^2+y_2^2+y_3^2\right)$$-2a\left(x_1+x_2+x_3\right)+3a^2$

Since $G$ is at origins so , $\begin{array}{l}\frac{x_1+x_2+x_3}{3}=0\\ \end{array}$

$\Rightarrow x_1+x_2+x_3=0$

$\Rightarrow LHS=\sum x_1^2+\sum y_1^2-0+3a^2$$=\sum x_1^2+\sum y_1^2+3a^2$

And $\mathrm{RHS}=G{A}^2+G{B}^2+G{C}^2+3G{O}^2$

$\Rightarrow RHS=x_1^2+y_1^2+x_2^2+y_2^2+x_3^2+y_3^2+3(a-0{)}^2$$=\sum x_1^2+\sum y_1^2+3a^2$

Thus we get $O{A}^2+O{B}^2+O{C}^2=G{A}^2+G{B}^2+G{C}^2+3G{O}^2$

Hence option-$1$ is the correct answer.