Let $$z \in C$$, the set of complex numbers. Then the equation, $$2|z + 3i| - |z - i| = 0$$ represents:

Let us write the complex number $$z$$ in the Cartesian form $$z=x+iy,$$ where $$x$$ and $$y$$ are real numbers representing the coordinates of the point in the Argand plane.

We have the given equation

$$2\lvert z+3i\rvert-\lvert z-i\rvert=0.$$

First, recall the definition: for a complex number $$u=a+ib,$$ its modulus is given by the formula $$\lvert u\rvert=\sqrt{a^{2}+b^{2}}.$$

Now we write each modulus explicitly:

1. $$z+3i = x+i(y+3) \quad\Rightarrow\quad \lvert z+3i\rvert = \sqrt{x^{2}+(y+3)^{2}}.$$

2. $$z-i = x+i(y-1) \quad\Rightarrow\quad \lvert z-i\rvert = \sqrt{x^{2}+(y-1)^{2}}.$$

Substituting these into the original equation gives

$$2\sqrt{x^{2}+(y+3)^{2}}-\sqrt{x^{2}+(y-1)^{2}}=0.$$

We isolate one of the square roots:

$$2\sqrt{x^{2}+(y+3)^{2}}=\sqrt{x^{2}+(y-1)^{2}}.$$

To remove the square roots, we square both sides (using the fact that both sides are non-negative):

$$\left[2\sqrt{x^{2}+(y+3)^{2}}\right]^{2}=\left[\sqrt{x^{2}+(y-1)^{2}}\right]^{2}.$$

Thus

$$4\bigl(x^{2}+(y+3)^{2}\bigr)=x^{2}+(y-1)^{2}.$$

Next, we expand every squared term:

$$4\Bigl(x^{2}+y^{2}+6y+9\Bigr)=x^{2}+y^{2}-2y+1.$$

So we have

$$4x^{2}+4y^{2}+24y+36 = x^{2}+y^{2}-2y+1.$$

Now we bring all terms to the left-hand side:

$$4x^{2}-x^{2}+4y^{2}-y^{2}+24y+2y+36-1=0.$$

This simplifies to

$$3x^{2}+3y^{2}+26y+35=0.$$

Dividing every term by $$3$$ for simplicity, we obtain

$$x^{2}+y^{2}+\frac{26}{3}y+\frac{35}{3}=0.$$

To recognize the locus, we complete the square in $$y$$. For the expression $$y^{2}+\frac{26}{3}y,$$ half the coefficient of $$y$$ is $$\tfrac{13}{3},$$ and its square is $$\left(\tfrac{13}{3}\right)^{2}=\tfrac{169}{9}.$$ We add and subtract this square inside the equation:

$$x^{2}+\Bigl[y^{2}+\tfrac{26}{3}y+\tfrac{169}{9}\Bigr]-\tfrac{169}{9}+\tfrac{35}{3}=0.$$

Combine the constant terms. Since $$\tfrac{35}{3}=\tfrac{105}{9},$$ we have

$$-\tfrac{169}{9}+\tfrac{105}{9}=-\tfrac{64}{9}.$$

Thus the equation becomes

$$x^{2}+\bigl(y+\tfrac{13}{3}\bigr)^{2}-\tfrac{64}{9}=0.$$

Moving the constant to the other side gives

$$x^{2}+\bigl(y+\tfrac{13}{3}\bigr)^{2}=\tfrac{64}{9}.$$

This is clearly the equation of a circle. Its centre is $$\bigl(0,\;-\tfrac{13}{3}\bigr)$$ and its radius is

$$\sqrt{\tfrac{64}{9}}=\tfrac{8}{3}.$$

Hence the locus represented by the given equation is a circle of radius $$\tfrac{8}{3}$$.

Hence, the correct answer is Option A.