Let $$\mathbb{R}$$ denote the set of all real numbers. Let $$z_1 = 1 + 2i$$ and $$z_2 = 3i$$ be two complex numbers, where $$i = \sqrt{-1}$$. Let

$$S = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x + iy - z_1| = 2|x + iy - z_2|\}.$$

Then which of the following statements is (are) TRUE?

Let a general point in the plane be $$P(x,y)$$, so that the complex number corresponding to $$P$$ is $$x+iy$$.

Given $$z_1 = 1+2i$$ and $$z_2 = 3i = 0+3i$$, the set $$S$$ is defined by
$$|x + iy - z_1| = 2\,|x + iy - z_2|.$$

Convert each modulus to distance form:
$$|x + iy - z_1| = |(x-1) + i(y-2)| = \sqrt{(x-1)^2 + (y-2)^2},$$
$$|x + iy - z_2| = |x + i(y-3)| = \sqrt{x^2 + (y-3)^2}.$$

Hence the locus condition is
$$\sqrt{(x-1)^2 + (y-2)^2} = 2\sqrt{x^2 + (y-3)^2}.$$

Square both sides to remove the square roots:
$$(x-1)^2 + (y-2)^2 = 4\bigl(x^2 + (y-3)^2\bigr).$$

Expand each side:
Left side: $$x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 - 2x - 4y + 5.$$
Right side: $$4\bigl(x^2 + y^2 - 6y + 9\bigr) = 4x^2 + 4y^2 - 24y + 36.$$

Move everything to the right side and combine like terms:
$$0 = 4x^2 + 4y^2 - 24y + 36 - (x^2 + y^2 - 2x - 4y + 5).$$
$$0 = 3x^2 + 3y^2 + 2x - 20y + 31.$$

Divide by $$3$$ to simplify:
$$x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0.$$

Complete the square in both $$x$$ and $$y$$.

For $$x$$:
$$x^2 + \frac{2}{3}x = \left(x + \frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2.$$

For $$y$$:
$$y^2 - \frac{20}{3}y = \left(y - \frac{10}{3}\right)^2 - \left(\frac{10}{3}\right)^2.$$

Substitute these back:
$$\left(x + \frac{1}{3}\right)^2 - \frac{1}{9} + \left(y - \frac{10}{3}\right)^2 - \frac{100}{9} + \frac{31}{3} = 0.$$

Convert all constants to a common denominator $$9$$:
$$-\frac{1}{9} - \frac{100}{9} + \frac{31}{3} = -\frac{101}{9} + \frac{93}{9} = -\frac{8}{9}.$$

Thus the equation of the locus becomes
$$\left(x + \frac{1}{3}\right)^2 + \left(y - \frac{10}{3}\right)^2 = \frac{8}{9}.$$

This is the standard form of a circle with

Centre : $$\left(-\frac{1}{3},\,\frac{10}{3}\right),$$
Radius : $$\sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}.$$

Comparing with the given options:

Option A (centre $$(-\frac{1}{3},\frac{10}{3})$$) is TRUE.
Option D (radius $$\frac{2\sqrt{2}}{3}$$) is TRUE.
Options B and C are false.

Hence, the correct statements are:
Option A and Option D.