Three parallel lines $$L_1,L_2,L_3$$ are drawn in the plane such that the perpendicular distance between $$L_1$$ and $$L_2$$ is $$3$$ and the perpendicular distance between $$L_2$$ and $$L_3$$ is also $$3$$. A square $$ABCD$$ is constructed such that $$A$$ lies on $$L_1$$, $$B$$ lies on $$L_3$$ and $$C$$ lies on $$L_2$$. Find the area of the square.

Let the three parallel lines be taken as the vertical lines
$$L_1 : x = 0,\;L_2 : x = 3,\;L_3 : x = 6,$$
so that the perpendicular distance between consecutive lines is $$3$$ units.

Place the vertices of the square as
$$A = (0,\,y_A)\in L_1,\qquad B = (6,\,y_B)\in L_3.$$

The side $$AB$$ of the square has vector
$$\vec{AB} = (6,\;y_B-y_A) = (6,\;\Delta y).$$
Its length (the side of the square) is therefore
$$s = \sqrt{6^{2}+(\Delta y)^{2}}.$$

To reach the next vertex $$C$$ of the square we must rotate $$\vec{AB}$$ by $$90^\circ$$, preserving length. A counter-clockwise rotation gives
$$\vec{BC} = (-\Delta y,\;6).$$
Hence
$$C = B + \vec{BC} = (6-\Delta y,\;y_B+6).$$

The point $$C$$ must lie on $$L_2$$, i.e. its $$x$$-coordinate must be $$3$$:

$$6 - \Delta y = 3 \;\Longrightarrow\; \Delta y = 3.$$

Thus
$$y_B = y_A + 3,\qquad s = \sqrt{6^{2}+3^{2}} = \sqrt{45}=3\sqrt{5}.$$

The area of the square is
$$\text{Area} = s^{2} = (3\sqrt{5})^{2} = 45.$$

If the square were built using the clockwise rotation of $$\vec{AB}$$, the calculation would give $$\Delta y = -3$$ and the same side length $$3\sqrt{5}$$, so the area is unchanged.

Hence, the required area of the square is $$45$$.