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

Let $$z(\theta)=\frac{1967+1686\,i\sin\theta}{\,7-3\,i\cos\theta\,},\qquad\theta\in\mathbb{R}.$$

Write $$\sin\theta=s,\;\cos\theta=c$$ with $$s^{2}+c^{2}=1.$$ To separate the real and imaginary parts, multiply numerator and denominator by the complex conjugate of the denominator:

$$z=\frac{1967+1686\,i s}{7-3\,i c}\times\frac{7+3\,i c}{7+3\,i c}=\frac{(1967+1686\,i s)(7+3\,i c)}{7^{2}+(3c)^{2}}.$$

Denominator:
$$7^{2}+(3c)^{2}=49+9c^{2}.$$

Numerator:
$$$ \begin{aligned} (1967+1686\,i s)(7+3\,i c)&=1967\!\times\!7+1967\!\times\!3\,i c+1686\,i s\!\times\!7+1686\,i s\!\times\!3\,i c\\ &=13769+5901\,i c+11802\,i s-5058\,s c. \end{aligned} $$$

Thus
$$z=\frac{\,13769-5058\,s c+i(5901\,c+11802\,s)\,}{49+9c^{2}}.$$

For $$z$$ to be a real number, its imaginary part must vanish:

$$5901\,c+11802\,s=0\;\;\Longrightarrow\;\;c=-2s.$$

Using $$s^{2}+c^{2}=1$$ gives
$$s^{2}+4s^{2}=5s^{2}=1\;\;\Longrightarrow\;\;s=\pm\frac1{\sqrt5},\;\;c=\mp\frac{2}{\sqrt5}.$$

The product $$s c=\left(\pm\frac1{\sqrt5}\right)\!\left(\mp\frac{2}{\sqrt5}\right)=-\frac{2}{5},$$ independent of the sign choice.

Compute denominator:
$$49+9c^{2}=49+9\left(\frac{4}{5}\right)=49+\frac{36}{5}=\frac{281}{5}.$$

Compute the real part of the numerator:
$$$ \begin{aligned} 13769-5058\,s c&=13769-5058\!\left(-\frac{2}{5}\right)\\ &=13769+\frac{10116}{5}=\frac{68845+10116}{5}=\frac{78961}{5}. \end{aligned} $$$

Therefore
$$z=\frac{\dfrac{78961}{5}}{\dfrac{281}{5}}=\frac{78961}{281}.$$

Since $$281^{2}=78961,$$ we get $$z=281.$

This is the only positive integer value attained, so $$n=281.$$

Answer: 281