Let $A=\left\{\frac{1967+1686i \sin \theta }{7-3i \cos \theta }: \theta \in R\right\}$. If $A$ contains exactly one positive integer $n$, then the value of $n$ is

Let $z=\frac{1967+1686i \sin \theta }{7-3i \cos \theta }$ is a positive integer.

$\Rightarrow z=\frac{\left(1967+1686i \sin \theta \right)\left(7+3i \cos \theta \right)}{\left(7-3i \cos \theta \right)\left(7+3i \cos \theta \right)}$

$\Rightarrow z=\frac{1967\times 7-1686\times 3 \sin \theta \cos \theta +i\left(1686\times 7 \sin \theta +1967\times 3 \cos \theta \right)}{49+9 {\cos }^2\theta }$

Now taking imaginary part as zero we get,

$1686\times 7 \sin \theta +1967\times 3 \cos \theta =0$

$\Rightarrow \left(281\times 6\right)\times 7 \sin \theta +\left(281\times 7\right)\times 3 \cos \theta =0$

$\Rightarrow 42\sin \theta +21\cos \theta =0$

$\Rightarrow \tan \theta =-\frac{1}{2}$

$\Rightarrow {\cos }^2\theta =\frac{4}{5} \& \sin \theta \cos \theta =-\frac{2}{5}$

Now putting the value in $z$we get,

$z=\frac{\left(281\times 7\times 7\right)-\left(281\times 6\right)\times 3\times \left(-\frac{2}{5}\right)}{49+9\times \frac{4}{5}}$

$\Rightarrow z=\frac{281\left(49+\frac{36}{5}\right)}{\left(49+\frac{36}{5}\right)}=281$

Hence, the value of $n$ is $281$