The distance between the two points $$A$$ and $$A'$$ which lie on $$y = 2$$ such that both the line segments $$AB$$ and $$A'B$$ (where $$B$$ is the point $$(2, 3)$$) subtend angle $$\frac{\pi}{4}$$ at the origin, is equal to

Points $$A$$ and $$A'$$ lie on $$y = 2$$, so let $$A = (a, 2)$$ and $$A' = (a', 2)$$. Given $$B = (2, 3)$$.

The line segments $$OA$$ and $$OB$$ subtend angle $$\frac{\pi}{4}$$ at the origin, meaning $$\angle AOB = \frac{\pi}{4}$$.

Slope of $$OB = \frac{3}{2}$$ and slope of $$OA = \frac{2}{a}$$.

Using the tangent formula for angle between two lines:

$$\tan\frac{\pi}{4} = \left|\frac{\frac{2}{a} - \frac{3}{2}}{1 + \frac{2}{a} \cdot \frac{3}{2}}\right| = \left|\frac{\frac{4 - 3a}{2a}}{\frac{a + 3}{a}}\right| = \left|\frac{4 - 3a}{2(a + 3)}\right| = 1$$

Case 1: $$4 - 3a = 2(a + 3) = 2a + 6$$

$$4 - 3a = 2a + 6 \implies -5a = 2 \implies a = -\frac{2}{5}$$

Case 2: $$4 - 3a = -(2a + 6) = -2a - 6$$

$$4 - 3a = -2a - 6 \implies -a = -10 \implies a = 10$$

So the two points are $$A = \left(-\frac{2}{5}, 2\right)$$ and $$A' = (10, 2)$$.

The distance between them is:

$$|AA'| = \left|10 - \left(-\frac{2}{5}\right)\right| = 10 + \frac{2}{5} = \frac{52}{5}$$

The answer is Option C: $$\dfrac{52}{5}$$.