Let $$S_1 = \{z_1 \in \mathbb{C} : |z_1 - 3| = \frac{1}{2}\}$$ and $$S_2 = \{z_2 \in \mathbb{C} : |z_2 - |z_2 + 1|| = |z_2 + |z_2 - 1||\}$$. Then, for $$z_1 \in S_1$$ and $$z_2 \in S_2$$, the least value of $$|z_2 - z_1|$$ is

We need to find the least value of $$|z_2 - z_1|$$ for $$z_1 \in S_1$$ and $$z_2 \in S_2$$.

Since $$S_1 = \{z_1 \in \mathbb{C} : |z_1 - 3| = \frac{1}{2}\}$$, it is a circle centred at $$(3, 0)$$ with radius $$\frac{1}{2}$$.

We have $$S_2 = \{z_2 \in \mathbb{C} : |z_2 - |z_2 + 1|| = |z_2 + |z_2 - 1||\}$$. Since $$|z_2 + 1|$$ and $$|z_2 - 1|$$ are non-negative real numbers, we set $$r_1 = |z_2 + 1|$$ and $$r_2 = |z_2 - 1|$$, and the condition becomes $$|z_2 - r_1| = |z_2 + r_2|$$.

Squaring both sides (valid since both sides are non-negative) gives $$|z_2 - r_1|^2 = |z_2 + r_2|^2$$. Letting $$z_2 = x + iy$$ leads to $$(x - r_1)^2 + y^2 = (x + r_2)^2 + y^2$$, which simplifies to $$-2xr_1 + r_1^2 = 2xr_2 + r_2^2$$. Equivalently, $$(r_1 + r_2)(r_1 - r_2) = 2x(r_1 + r_2)$$.

Since $$r_1 + r_2 = |z_2 + 1| + |z_2 - 1| > 0$$, dividing by $$(r_1 + r_2)$$ yields $$r_1 - r_2 = 2x$$, namely $$|z_2 + 1| - |z_2 - 1| = 2\text{Re}(z_2) \quad \cdots (*)$$.

Substituting $$z_2 = x + iy$$ into (*) gives $$\sqrt{(x+1)^2 + y^2} - \sqrt{(x-1)^2 + y^2} = 2x$$. Rearranging leads to $$\sqrt{(x+1)^2 + y^2} = 2x + \sqrt{(x-1)^2 + y^2}$$, requiring $$2x + \sqrt{(x-1)^2 + y^2} \geq 0$$, which holds for all $$x \geq 0$$ and also for some $$x < 0$$.

Squaring both sides again yields $$(x+1)^2 + y^2 = 4x^2 + 4x\sqrt{(x-1)^2 + y^2} + (x-1)^2 + y^2$$, which simplifies through cancellation to $$4x = 4x^2 + 4x\sqrt{(x-1)^2 + y^2}$$ and hence $$4x\left(1 - x - \sqrt{(x-1)^2 + y^2}\right) = 0$$.

One possibility is $$x = 0$$. Substituting back into (*) gives $$|iy + 1| - |iy - 1| = 0$$, and since $$\sqrt{1+y^2} = \sqrt{1+y^2}$$ holds identically, all points on the imaginary axis are in $$S_2$$.

The other possibility with $$x \neq 0$$ forces $$1 - x = \sqrt{(x-1)^2 + y^2}$$, which requires $$1 - x \geq 0$$ (so $$x \leq 1$$), and squaring gives $$(1-x)^2 = (x-1)^2 + y^2$$, hence $$y^2 = 0$$ or $$y = 0$$. For $$y = 0$$ and $$-1 \leq x \leq 1$$, one checks that $$|x+1| - |x-1| = (x+1) - (1-x) = 2x = 2x$$, so these points satisfy (*). For $$x > 1$$ or $$x < -1$$ the equality fails except at the endpoints $$x = \pm 1$$.

Therefore, $$S_2$$ consists of the entire imaginary axis together with the real line segment $$[-1, 1]$$.

Finally, since the circle $$S_1$$ is centred at $$(3, 0)$$ with radius $$\frac{1}{2}$$, the closest part of $$S_2$$ to this circle is the point $$(1, 0)$$ on the real segment. The point on $$S_1$$ nearest to $$(1, 0)$$ lies in the same direction and is $$\left(3 - \frac{1}{2}, 0\right) = \left(\frac{5}{2}, 0\right)$$, giving a minimum distance of $$\frac{5}{2} - 1 = \frac{3}{2}$$.

Therefore, the least value of $$|z_2 - z_1|$$ is $$\frac{3}{2}$$.

The correct answer is Option C: $$\frac{3}{2}$$.