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

S=\left\{\right(x,y)\in {ℝ}^2:|x+iy-z_1|=2|x+iy-z_2\left|\right\}

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

To determine the properties of the set $S$, we start by representing the complex equation in Cartesian coordinates.

Let $z=x+iy$ be a complex number where $x,y\in ℝ$.
The given complex numbers are:
$z_1=1+2i$
$z_2=3i$

The equation defining the set $S$ is given by:
$|z-z_1|=2|z-z_2|$

Squaring both sides of the equation yields:
${|z-z_1|}^2=4{|z-z_2|}^2$

Now, we substitute the expressions for $z-z_1$ and $z-z_2$:
$z-z_1=(x-1)+i(y-2)$
$z-z_2=x+i(y-3)$

Using the definition of the modulus of a complex number, ${|a+ib|}^2=a^2+b^2$, we write:
${(x-1)}^2+{(y-2)}^2=4[x^2+{(y-3)}^2]$

Expanding the algebraic terms on both sides:
$(x^2-2x+1)+(y^2-4y+4)=4[x^2+(y^2-6y+9\left)\right]$
$x^2+y^2-2x-4y+5=4x^2+4y^2-24y+36$

Rearranging all terms to one side of the equation:
$3x^2+3y^2+2x-20y+31=0$

Dividing the entire equation by 3 to bring it into the standard form of a circle's equation:
$x^2+y^2+\frac{2}{3}x-\frac{20}{3}y+\frac{31}{3}=0$

Comparing this with the general equation of a circle, $x^2+y^2+2gx+2fy+c=0$, we identify:
$2g=\frac{2}{3}\Rightarrow g=\frac{1}{3}$
$2f=-\frac{20}{3}\Rightarrow f=-\frac{10}{3}$
$c=\frac{31}{3}$

The centre of the circle is given by $(-g,-f)$:
$\text{Centre}=(-\frac{1}{3},\frac{10}{3})$

The radius $R$ of the circle is computed using the formula $R=\sqrt{g^2+f^2-c}$:
$R=\sqrt{{\left(\frac{1}{3}\right)}^2+{(-\frac{10}{3})}^2-\frac{31}{3}}$
$R=\sqrt{\frac{1}{9}+\frac{100}{9}-\frac{93}{9}}$
$R=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}$

Therefore, the set $S$ represents a circle with centre $(-\frac{1}{3},\frac{10}{3})$ and radius $\frac{2\sqrt{2}}{3}$.