Let $$B$$ and $$C$$ be the two points on the line $$y + x = 0$$ such that $$B$$ and $$C$$ are symmetric with respect to the origin. Suppose $$A$$ is a point on $$y - 2x = 2$$ such that $$\triangle ABC$$ is an equilateral triangle. Then, the area of the $$\triangle ABC$$ is

Since $$B$$ and $$C$$ lie on the line $$x + y = 0$$ and are symmetric with respect to the origin, the origin $$O(0, 0)$$ is the midpoint of the segment $$BC$$. In an equilateral triangle $$\triangle ABC$$, the median from vertex $$A$$ to the base $$BC$$ is also the altitude. Therefore, the line $$AO$$ must be perpendicular to the line $$x + y = 0$$.

The slope of the line $$x + y = 0$$ is $$-1$$. Since $$AO$$ is perpendicular to it, the slope of the line $$AO$$ is $$1$$. Since $$AO$$ passes through the origin $$(0, 0)$$, its equation is:

$$y = x$$

The vertex $$A$$ is given to lie on the line $$y - 2x = 2$$. To find the coordinates of $$A$$, we solve the intersection of $$y = x$$ and $$y - 2x = 2$$ by substitution:

$$x - 2x = 2$$

$$-x = 2 \implies x = -2$$

$$y = -2$$

Thus, the coordinates of vertex $$A$$ are $$(-2, -2)$$.

The length of the altitude $$h$$ of the equilateral triangle is the distance from $$A(-2, -2)$$ to the origin $$O(0, 0)$$:

$$h = \sqrt{(-2 - 0)^2 + (-2 - 0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$$

The area $$\Delta$$ of an equilateral triangle in terms of its altitude $$h$$ is given by the formula:

$$\Delta = \frac{h^2}{\sqrt{3}}$$

Substituting the value of $$h$$ into the area formula:

$$\Delta = \frac{(2\sqrt{2})^2}{\sqrt{3}} = \frac{8}{\sqrt{3}}$$

Conclusion:

The area of the $$\triangle ABC$$ is equal to $$\frac{8}{\sqrt{3}}$$.