Let $$PQ$$ be a chord of the parabola $$y^2 = 12x$$ and the midpoint of $$PQ$$ be at $$(4, 1)$$. Then, which of the following point lies on the line passing through the points $$P$$ and $$Q$$?

For parabola $$y^2 = 4ax$$ (here $$4a = 12$$, so $$a = 3$$), the equation of chord with midpoint $$(h, k)$$ is: $$T = S_1$$ where $$T: ky - 2a(x+h) = 0$$ and $$S_1 = k^2 - 4ah$$.

Substituting $$h=4$$ and $$k=1$$ into $$T$$ gives $$T: 1 \cdot y - 6(x+4) = k^2 - 4ah$$ and hence $$y - 6x - 24 = 1 - 48 = -47$$. Since $$S_1 = k^2 - 12h = 1 - 48 = -47$$, equating $$T = S_1$$ yields $$ky - 6(x+h) = k^2 - 12h$$, so $$1 \cdot y - 6(x+4) = 1 - 48$$ which simplifies to $$y - 6x - 24 = -47$$.

Rearranging gives $$y - 6x = -23$$ and thus $$y = 6x - 23$$.

Testing Option A $$(3, -3)$$ leads to $$-3 = 18-23 = -5$$, so it does not lie on the line. Option B $$(2, -9)$$ yields $$-9 = 12-23 = -11$$, and Option C $$(3/2, -16)$$ yields $$-16 = 9-23 = -14$$, both of which fail. However, Option D $$(1/2, -20)$$ gives $$-20 = 3-23 = -20$$.

Therefore, the point $$\left(\frac{1}{2}, -20\right)$$ lies on the line, which matches Option D, so the answer is Option D.