For $\mathrm{n}\in \mathrm{N}$, let $S_n=\left\{z\in C:|z-3+2i|=\frac{n}{4}\right\}$ and ${\mathrm{T}}_{\mathrm{n}}=\left\{\mathrm{z}\in \mathrm{C}:|\mathrm{z}-2+3\mathrm{i}|=\frac{1}{\mathrm{n}}\right\}$. Then the number of elements in the set $\left\{\mathrm{n}\in \mathrm{N}:{\mathrm{S}}_{\mathrm{n}}\cap {\mathrm{T}}_{\mathrm{n}}=\varphi \right\}$ is:

$S_n:|z-(3-2i\left)\right|=\frac{n}{4}$ is a circle center ${\mathrm{C}}_1(3,-2)$ and radius n/4

$T_n:|z-(2-3i\left)\right|=\frac{1}{n}$is a circle center ${\mathrm{C}}_2(2,-3)$and radius 1/n

Here $S_n\cap T_n=\varphi$

Both circles do not intersect each other

Case- 1: ${\mathrm{C}}_1{\mathrm{C}}_2>\mathrm{n}/4+1/\mathrm{n}$

$\sqrt{2}>\frac{n}{4}+\frac{1}{n}$

then n = 1, 2, 3, 4

Case- 2:${\mathrm{C}}_1{\mathrm{C}}_2<\left|\frac{\mathrm{n}}{4}-\frac{1}{\mathrm{n}}\right|$

$\Rightarrow \sqrt{2}<\left|\frac{{\mathrm{n}}^2-4}{4\mathrm{n}}\right|$

$\Rightarrow$n has infinite solutions for$\mathrm{n}\in \mathrm{N}$

note: Correct answer  is infinite solutions  but answer given by NTA is 4