If $$z$$ is a complex number such that $$|z| \leq 1$$, then the minimum value of $$\left|z + \frac{1}{2}(3 + 4i)\right|$$ is:

$$\frac{1}{2}(3 + 4i) = \frac{3}{2} + 2i$$

So the expression becomes $$\left|z + \left(\frac{3}{2} + 2i\right)\right|$$.

This can be rewritten as $$\left|z - \left(-\frac{3}{2} - 2i\right)\right|$$, which represents the distance between $$z$$ and the point $$P\left(-\frac{3}{2}, -2\right)$$ in the complex plane.

The condition $$|z| \leq 1$$ describes a closed disk of radius 1 centered at the origin $$(0, 0)$$.

To find the minimum distance from any point in this disk to the point $$P\left(-\frac{3}{2}, -2\right)$$, first compute the distance from the center $$(0, 0)$$ to $$P$$:

$$\left|-\frac{3}{2} - 2i\right| = \sqrt{\left(-\frac{3}{2}\right)^2 + (-2)^2} = \sqrt{\frac{9}{4} + 4} = \sqrt{\frac{9}{4} + \frac{16}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$

Since $$\frac{5}{2} = 2.5 > 1$$ (the radius of the disk), the point $$P$$ lies outside the disk.

The minimum distance from a point in the disk to an external point $$P$$ is the distance from the center to $$P$$ minus the radius of the disk:

$$\text{Minimum distance} = \frac{5}{2} - 1 = \frac{5}{2} - \frac{2}{2} = \frac{3}{2}$$

This minimum value is achieved when $$z$$ lies on the line segment joining the origin to $$P$$ and on the boundary of the disk (i.e., $$|z| = 1$$). Specifically, $$z$$ is in the direction opposite to $$P$$:

The vector from $$(0, 0)$$ to $$P$$ is $$\left(-\frac{3}{2}, -2\right)$$. The unit vector in this direction is:

$$\left( \frac{-\frac{3}{2}}{\frac{5}{2}}, \frac{-2}{\frac{5}{2}} \right) = \left( -\frac{3}{2} \cdot \frac{2}{5}, -2 \cdot \frac{2}{5} \right) = \left( -\frac{3}{5}, -\frac{4}{5} \right)$$

So $$z = -\frac{3}{5} - \frac{4}{5}i$$

The correct option is C.