Let the circle $$S : 36x^2 + 36y^2 - 108x + 120y + C = 0$$ be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, $$x - 2y = 4$$ and $$2x - y = 5$$ lies inside the circle $$S$$, then:

First, divide the equation of circle $$S$$ by 36 to get standard form: $$x^2 + y^2 - 3x + \frac{10}{3}y + \frac{C}{36} = 0$$.

The centre is $$\left(\frac{3}{2}, -\frac{5}{3}\right)$$ and the radius is $$r = \sqrt{\frac{9}{4} + \frac{25}{9} - \frac{C}{36}} = \sqrt{\frac{81 + 100 - C}{36}} = \frac{\sqrt{181 - C}}{6}$$.

For the circle to exist, we need $$181 - C > 0$$, i.e., $$C < 181$$.

The centre is at $$\left(\frac{3}{2}, -\frac{5}{3}\right)$$, which is in the fourth quadrant.

Condition 1 (no touch/intersection with x-axis): The distance from the centre to the $$x$$-axis is $$\left|-\frac{5}{3}\right| = \frac{5}{3}$$. We need $$r < \frac{5}{3}$$, so $$\frac{\sqrt{181-C}}{6} < \frac{5}{3}$$, giving $$\sqrt{181-C} < 10$$, so $$181 - C < 100$$, i.e., $$C > 81$$.

Condition 2 (no touch/intersection with y-axis): The distance from the centre to the $$y$$-axis is $$\frac{3}{2}$$. We need $$r < \frac{3}{2}$$, so $$\frac{\sqrt{181-C}}{6} < \frac{3}{2}$$, giving $$\sqrt{181-C} < 9$$, so $$181 - C < 81$$, i.e., $$C > 100$$.

The stricter condition is $$C > 100$$.

Now find the intersection of $$x - 2y = 4$$ and $$2x - y = 5$$. From the first equation, $$x = 4 + 2y$$. Substituting: $$2(4 + 2y) - y = 5$$, so $$8 + 3y = 5$$, giving $$y = -1$$ and $$x = 2$$. The point is $$(2, -1)$$.

This point must lie inside the circle: distance from $$(2, -1)$$ to centre $$\left(\frac{3}{2}, -\frac{5}{3}\right)$$ must be less than $$r$$.

Distance squared: $$\left(\frac{1}{2}\right)^2 + \left(\frac{2}{3}\right)^2 = \frac{1}{4} + \frac{4}{9} = \frac{9 + 16}{36} = \frac{25}{36}$$.

We need $$\frac{25}{36} < r^2 = \frac{181-C}{36}$$, so $$25 < 181 - C$$, giving $$C < 156$$.

Combining: $$100 < C < 156$$.

The answer is $$100 < C < 156$$, which is Option D.