If $x_1,x_2,x_3$ as well as $y_1,y_2,y_3$ are in $GP$, with same common ratio, then the points $\left(x_1,y_1\right),\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$

Let common ratio of GP is $r$, then,

 ${\mathrm{x}}_2={\mathrm{x}}_1\mathrm{r} , {\mathrm{x}}_3={\mathrm{x}}_1{\mathrm{r}}^2$ and ${\mathrm{y}}_2={\mathrm{y}}_1\mathrm{r} , {\mathrm{y}}_3={\mathrm{y}}_1{\mathrm{r}}^2$ 

Let $\mathrm{A}\equiv \left({\mathrm{x}}_1,{\mathrm{y}}_1\right),\mathrm{B}\equiv \left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ and $\mathrm{C}\equiv \left({\mathrm{x}}_3,{\mathrm{y}}_3\right)$

$\therefore$As we know that Area of $∆\mathrm{ABC}=\frac{1}{2}\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_2& {\mathrm{y}}_2& 1\\ {\mathrm{x}}_3& {\mathrm{y}}_3& 1\end{array}\right|$

Thus, Area of $∆\mathrm{ABC}=\frac{1}{2}\left|\begin{array}{ccc}{\mathrm{x}}_1& {\mathrm{y}}_1& 1\\ {\mathrm{x}}_1\mathrm{r}& {\mathrm{y}}_1\mathrm{r}& 1\\ {\mathrm{x}}_1{\mathrm{r}}^2& {\mathrm{y}}_1{\mathrm{r}}^2& 1\end{array}\right|=\frac{x_1{y}_1}{2}\left|\begin{array}{ccc}1& 1& 1\\ \mathrm{r}& \mathrm{r}& 1\\ {\mathrm{r}}^2& {\mathrm{r}}^2& 1\end{array}\right|$

Since two columns of the determinant is same so $\left|\begin{array}{ccc}1& 1& 1\\ \mathrm{r}& \mathrm{r}& 1\\ {\mathrm{r}}^2& {\mathrm{r}}^2& 1\end{array}\right|=0$.

$\Rightarrow$Area of $△ABC=0$

Thus, the three points are collinear or lie on a straight line.

Hence option-$1$ is the correct answer.