Let $$A$$ be the point (1, 2) and $$B$$ be any point on the curve $$x^2 + y^2 = 16$$. If the centre of the locus of the point $$P$$, which divides the line segment AB in the ratio 3:2 is the point $$C(\alpha, \beta)$$, then the length of the line segment $$AC$$ is

We need to find the centre $$C(\alpha,\beta)$$ of the locus of point P that divides AB in the ratio 3:2 where A(1,2) and B lies on $$x^2 + y^2 = 16$$, and then determine the length of the line segment AC.

Since B lies on the circle $$x^2 + y^2 = 16$$ of radius 4 centered at the origin, we may parametrize it as $$B = (4\cos\theta, 4\sin\theta)$$. The point P divides AB in the ratio 3:2 from A to B, so by the section formula
$$P = \frac{2A + 3B}{5} = \left(\frac{2 + 12\cos\theta}{5}, \frac{4 + 12\sin\theta}{5}\right).$$

Writing $$P = (h,k)$$ leads to
$$h = \frac{2 + 12\cos\theta}{5}\;\implies\;\cos\theta = \frac{5h - 2}{12},$$
$$k = \frac{4 + 12\sin\theta}{5}\;\implies\;\sin\theta = \frac{5k - 4}{12}.$$ Using the identity $$\cos^2\theta + \sin^2\theta = 1$$ gives
$$\left(\frac{5h-2}{12}\right)^2 + \left(\frac{5k-4}{12}\right)^2 = 1,$$ which simplifies to
$$(5h-2)^2 + (5k-4)^2 = 144.$$ This is the equation of a circle with centre determined by $$5h - 2 = 0$$ and $$5k - 4 = 0$$, namely
$$h = \frac{2}{5},\quad k = \frac{4}{5}.$$ Hence
$$C = \left(\frac{2}{5}, \frac{4}{5}\right).$$

The distance $$AC$$ is then
$$AC = \sqrt{\left(1 - \frac{2}{5}\right)^2 + \left(2 - \frac{4}{5}\right)^2} = \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{6}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{36}{25}} = \sqrt{\frac{45}{25}} = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}.$$

The correct answer is Option 1: $$\frac{3\sqrt{5}}{5}$$.