Let $$z$$ be a complex number such that $$\left|\frac{z-i}{z+2i}\right| = 1$$ and $$|z| = \frac{5}{2}$$. Then, the value of $$|z + 3i|$$ is:

We are given two conditions on the complex number $$z$$.

First, we know that $$\left|\dfrac{z-i}{\,z+2i\,}\right| = 1$$. For any two complex numbers $$a$$ and $$b$$ we have the property $$\left|\dfrac{a}{b}\right| = \dfrac{|a|}{|b|}$$. So the given relation immediately gives

$$\dfrac{|\,z-i\,|}{|\,z+2i\,|} = 1 \;\;\Longrightarrow\;\; |\,z-i\,| = |\,z+2i\,|.$$

Let us write $$z$$ in its Cartesian form $$z = x + iy$$, where $$x$$ and $$y$$ are real numbers.

Then $$z - i = x + i(y-1)$$ and $$z + 2i = x + i(y+2).$$

The modulus of a complex number $$u + iv$$ is $$\sqrt{u^{2}+v^{2}}$$, so

$$|\,z-i\,| = \sqrt{x^{2} + (y-1)^{2}},\qquad |\,z+2i\,| = \sqrt{x^{2} + (y+2)^{2}}.$$

Because the moduli are equal, we have

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

Both sides are non-negative, so we can square without changing the equality:

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

The $$x^{2}$$ terms cancel, leaving

$$(y-1)^{2} = (y+2)^{2}.$$

Expanding both squares,

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

Subtracting $$y^{2}$$ from both sides and gathering like terms,

$$-2y + 1 = 4y + 4 \;\;\Longrightarrow\;\; -6y = 3 \;\;\Longrightarrow\;\; y = -\dfrac{1}{2}.$$

So the imaginary part of $$z$$ is fixed at $$y = -\dfrac{1}{2}$$.

Secondly, we are told that $$|z| = \dfrac{5}{2}.$$ Using $$|x + iy| = \sqrt{x^{2} + y^{2}}$$, we get

$$\sqrt{x^{2} + y^{2}} = \dfrac{5}{2} \;\;\Longrightarrow\;\; x^{2} + y^{2} = \left(\dfrac{5}{2}\right)^{2} = \dfrac{25}{4}.$$

Substituting $$y = -\dfrac{1}{2}$$, we find

$$x^{2} + \left(-\dfrac{1}{2}\right)^{2} = \dfrac{25}{4} \;\;\Longrightarrow\;\; x^{2} + \dfrac{1}{4} = \dfrac{25}{4}.$$

Isolating $$x^{2}$$,

$$x^{2} = \dfrac{25}{4} - \dfrac{1}{4} = \dfrac{24}{4} = 6.$$

Thus $$x = \pm\sqrt{6}.$$ So the two possible numbers are $$z = \sqrt{6} - \dfrac{i}{2}$$ and $$z = -\sqrt{6} - \dfrac{i}{2}.$$

We now compute $$|\,z + 3i\,|$$. For either value of $$z$$ we have

$$z + 3i = x + iy + 3i = x + i\bigl(y + 3\bigr).$$

Because $$y = -\dfrac{1}{2}$$, we get

$$y + 3 = -\dfrac{1}{2} + 3 = \dfrac{5}{2},$$

so

$$z + 3i = x + i\left(\dfrac{5}{2}\right).$$

The modulus is therefore

$$|\,z + 3i\,| = \sqrt{x^{2} + \left(\dfrac{5}{2}\right)^{2}}.$$

Substituting $$x^{2} = 6$$,

$$|\,z + 3i\,| = \sqrt{6 + \dfrac{25}{4}} = \sqrt{\dfrac{24}{4} + \dfrac{25}{4}} = \sqrt{\dfrac{49}{4}} = \dfrac{7}{2}.$$

Both possible values of $$z$$ give the same modulus, so the required value is $$\dfrac{7}{2}$$.

Hence, the correct answer is Option B.