The distance of the point $$(6, -2\sqrt{2})$$ from the common tangent $$y = mx + c$$, $$m > 0$$, of the curves $$x = 2y^2$$ and $$x = 1 + y^2$$ is

We need to find the common tangent $$y = mx + c$$ with $$m > 0$$ to the curves $$x = 2y^2$$ and $$x = 1 + y^2$$.

For the curve $$x = 2y^2$$, at the point $$(2t^2, t)$$ the derivative is $$\frac{dy}{dx} = \frac{1}{4t}$$. Hence the tangent line can be written as $$y = \frac{x}{4t} + \frac{t}{2}$$, which shows that $$m = \frac{1}{4t}$$ and $$c = \frac{t}{2}$$.

For the curve $$x = 1 + y^2$$, at the point $$(1 + s^2, s)$$ the derivative is $$\frac{dy}{dx} = \frac{1}{2s}$$. The tangent line is then $$y = \frac{x}{2s} - \frac{1}{2s} + \frac{s}{2}$$, giving $$m = \frac{1}{2s}$$ and $$c = -\frac{1}{2s} + \frac{s}{2}$$.

In order for these tangents to coincide, their slopes must be equal, so $$\frac{1}{4t} = \frac{1}{2s}$$ which implies $$s = 2t$$. Their intercepts must also match, so

$$\frac{t}{2} = -\frac{1}{2s} + \frac{s}{2} = -\frac{1}{4t} + t$$

Rearranging gives $$-\frac{t}{2} = -\frac{1}{4t}$$, and hence $$2t^2 = 1$$, so $$t = \frac{1}{\sqrt{2}}$$.

Substituting back, we find $$m = \frac{1}{4t} = \frac{\sqrt{2}}{4}$$ and $$c = \frac{t}{2} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$$.

Thus the common tangent is $$y = \frac{\sqrt{2}}{4}x + \frac{\sqrt{2}}{4}$$, which can also be written as $$\sqrt{2}\,x - 4y + \sqrt{2} = 0$$.

The distance from the point $$(6, -2\sqrt{2})$$ to this line is

$$d = \frac{\bigl|\sqrt{2}\,(6) - 4(-2\sqrt{2}) + \sqrt{2}\bigr|}{\sqrt{2 + 16}} = \frac{|6\sqrt{2} + 8\sqrt{2} + \sqrt{2}|}{\sqrt{18}} = \frac{15\sqrt{2}}{3\sqrt{2}} = 5\,. $$

Therefore, the correct answer is Option B: $$5$$.