Sum of the distances of a point from two perpendicular lines is 3. The area enclosed by the locus of the point is

let P=(x, y) be locus of point.

Given that $\left|\mathrm{x}\right|+\left|\mathrm{y}\right|=3$

$\Rightarrow \pm \mathrm{x}\pm \mathrm{y}=3$

locus represents a square

$$\begin{gathered} From triangle OAB,O{A}^2+O{B}^2=A{B}^2 \\ \Rightarrow A{B}^2=18 \\ \Rightarrow AB=3\sqrt{2} \end{gathered}$$

$\therefore$ Area of square $={\left(3\sqrt{2}\right)}^2$

                                 $=18$