$\begin{aligned} & \quad O A=O B \\ & A(x, y) \equiv \\ & x=0+r \cos \theta \\ & y=0+r \sin \theta \\ & (x, y) \equiv(r \cos \theta, r \sin \theta)\end{aligned}$


$\begin{aligned} & \text { For } B, x=0+r \cos \left(\theta+90^{\circ}\right)=-r \sin \theta \\ & y=0+r \sin \left(90^{\circ}+\theta\right) \\ & y=r \cos \theta \\ & B(-r \sin \theta, r \cos \theta) \\ & \because \quad A, B \operatorname{lies} \text { on } 2 x+3 y=6 \\ & \therefore \quad 2 r \cos \theta+3 r \sin \theta=6 \\ & \quad 3 r \cos \theta-2 r \sin \theta=6 \\ & \therefore \quad 2 r \cos \theta+3 r \sin \theta=3 r \cos \theta-2 r \sin \theta \\ & \quad 5 r \sin \theta=r \cos \theta \\ & \quad \tan \theta=\frac{1}{5} \\ & \quad \sin \theta=\frac{1}{\sqrt{26}}, \cos \theta=\frac{5}{\sqrt{26}} \\ & \quad \frac{2 \times 5}{\sqrt{26}} r+\frac{3 \times 1}{\sqrt{26}} r=6 \\ & \Rightarrow \quad 13 r=6 \sqrt{26} \Rightarrow r=\frac{6 \sqrt{2}}{\sqrt{13}} \\ & \text { Area }=\frac{1}{2} \times r \times r=\frac{r^2}{2}=\frac{36 \times 2}{2 \times 13}=\frac{36}{13} .\end{aligned}$