Given, that Area of triangle $=18$
$\Rightarrow \, \frac{1}{2}\begin{vmatrix}2 a & a & 1 \\ a & 2 a & 1 \\ a & a & 1\end{vmatrix}=\pm 18$
$\Rightarrow \, \begin{vmatrix}2 a & a & 1 \\ a & 2 a & 1 \\ a & a & 1\end{vmatrix}=\pm 36$
$\Rightarrow 2 a(2 a-a)-a(a-a)+1\left(a^{2}-2 a^{2}\right)=\pm 36$
$\Rightarrow 2 a^{2}-a^{2}=\pm 36$
$\Rightarrow a^{2}=\pm 36$
$\Rightarrow a^{2}=36$
$\Rightarrow a=\pm 6$
Now, centroid of the given triangle will be
$=\left(\frac{2 a + a + a}{3}, \frac{a+2 a + a}{3}\right)=\left(\frac{4 a}{3}, \frac{4 a}{3}\right)$
When $a=6$, centroid $=\left(\frac{4 \times 6}{3}, \frac{4 \times 6}{3}\right)=(8,8)$