For n ∈ N , let S _ n = \ z ∈ C: |z - 3 + 2i| = (n)/(4)\ and T _ n...

Here ${S}_{n}: \left|z - \left(3 - 2i\right)\right| = \dfrac{n}{4}$ represents a circle with center ${C}_{1}\left(3, - 2\right)$ and radius $\dfrac{n}{4}$

and ${T}_{n}: \left|z - \left(2 - 3i\right)\right| = \dfrac{1}{n}$ represents a circle with center ${C}_{2}\left(2, - 3\right)$ and radius $\dfrac{1}{n}$

For ${S}_{n} \cap {T}_{n} = ϕ$, both circles do not intersect each other.

When ${C}_{1}{C}_{2} > \dfrac{n}{4} + \dfrac{1}{n}$

i.e. $\sqrt{2} > \dfrac{n}{4} + \dfrac{1}{n}$
then possible values of $n = 1, 2, 3, 4$

When ${C}_{1}{C}_{2} < \left|\dfrac{n}{4} - \dfrac{1}{n}\right|$

$⇒ \sqrt{2} < \left|\dfrac{{n}^{2} - 4}{4n}\right|$

then $n$ has infinite solutions for $n \in N$

Hence, there are total four values possible.