The distance of the point $$(2, 3)$$ from the line $$2x - 3y + 28 = 0$$, measured parallel to the line $$\sqrt{3}x - y + 1 = 0$$, is equal to

We need to find the distance of the point $$(2, 3)$$ from the line $$2x - 3y + 28 = 0$$, measured parallel to the line $$\sqrt{3}x - y + 1 = 0$$.

The line $$\sqrt{3}x - y + 1 = 0$$ has slope $$m = \sqrt{3}$$, corresponding to an angle $$\theta = 60°$$ with the positive x-axis, so its direction ratios are $$(\cos 60°, \sin 60°) = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$$.

Thus a line through $$(2, 3)$$ parallel to this direction can be written in parametric form as $$x = 2 + r\cos 60° = 2 + \frac{r}{2}$$ and $$y = 3 + r\sin 60° = 3 + \frac{r\sqrt{3}}{2}$$.

Substituting these into the equation $$2x - 3y + 28 = 0$$ gives $$2\left(2 + \frac{r}{2}\right) - 3\left(3 + \frac{r\sqrt{3}}{2}\right) + 28 = 0,$$ which simplifies to $$4 + r - 9 - \frac{3r\sqrt{3}}{2} + 28 = 0,$$ $$23 + r - \frac{3\sqrt{3}r}{2} = 0,$$ $$23 + r\left(1 - \frac{3\sqrt{3}}{2}\right) = 0,$$ $$r = \frac{-23}{1 - \frac{3\sqrt{3}}{2}} = \frac{-23}{\frac{2 - 3\sqrt{3}}{2}} = \frac{-46}{2 - 3\sqrt{3}}.$$

Rationalising the denominator, we get $$r = \frac{-46}{2 - 3\sqrt{3}} \times \frac{2 + 3\sqrt{3}}{2 + 3\sqrt{3}} = \frac{-46(2 + 3\sqrt{3})}{4 - 27} = \frac{-46(2 + 3\sqrt{3})}{-23},$$ $$r = 2(2 + 3\sqrt{3}) = 4 + 6\sqrt{3}.$$

The required distance is $$|r| = 4 + 6\sqrt{3}$$. Therefore, the correct answer is Option 4: $$4 + 6\sqrt{3}$$.