Let the lines $$(2 - i)z = (2 + i)\bar{z}$$ and $$(2 + i)z + (i - 2)\bar{z} - 4i = 0$$, (here $$i^2 = -1$$) be normal to a circle $$C$$. If the line $$iz + \bar{z} + 1 + i = 0$$ is tangent to this circle $$C$$, then its radius is:

We need to find the equations of the two normal lines and the tangent line by converting from complex number form to Cartesian form. Let $$z = x + iy$$ and $$\bar{z} = x - iy$$.

For Line 1: $$(2 - i)z = (2 + i)\bar{z}$$. Expanding: $$(2 - i)(x + iy) = (2 + i)(x - iy)$$. The left side gives $$2x + 2iy - ix - i^2y = 2x + y + i(2y - x)$$. The right side gives $$2x - 2iy + ix - i^2y = 2x + y + i(x - 2y)$$. Equating imaginary parts: $$2y - x = x - 2y$$, which gives $$4y = 2x$$, or $$x = 2y$$.

For Line 2: $$(2 + i)z + (i - 2)\bar{z} - 4i = 0$$. Expanding: $$(2 + i)(x + iy) + (i - 2)(x - iy) - 4i = 0$$. This gives $$(2x - y + ix + 2iy) + (-2x + y + ix + 2iy) - 4i = 0$$, which simplifies to $$2ix + 4iy - 4i = 0$$. Dividing by $$2i$$: $$x + 2y - 2 = 0$$, or $$x + 2y = 2$$.

The centre of the circle is the intersection of the two normals. From $$x = 2y$$ and $$x + 2y = 2$$: substituting gives $$2y + 2y = 2$$, so $$y = \frac{1}{2}$$ and $$x = 1$$. The centre is $$\left(1, \frac{1}{2}\right)$$.

For the tangent line: $$iz + \bar{z} + 1 + i = 0$$. Expanding: $$i(x + iy) + (x - iy) + 1 + i = 0$$, which gives $$(ix - y) + (x - iy) + 1 + i = 0$$. Separating real and imaginary parts: $$(x - y + 1) + i(x - y + 1) = 0$$. Both parts give $$x - y + 1 = 0$$.

The radius equals the perpendicular distance from the centre $$\left(1, \frac{1}{2}\right)$$ to the tangent line $$x - y + 1 = 0$$: $$r = \frac{|1 - \frac{1}{2} + 1|}{\sqrt{1^2 + (-1)^2}} = \frac{\frac{3}{2}}{\sqrt{2}} = \frac{3}{2\sqrt{2}}$$

Therefore, the radius of the circle is $$\dfrac{3}{2\sqrt{2}}$$, which corresponds to Option 3.