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

$\begin{array}{l}z=\frac{1967+1686i\sin \theta }{7-3i\cos \theta }\\ =\frac{(1967+168i\sin \theta )\times (7+3i\cos \theta )}{49+9{\cos }^2\theta }\\ \therefore \mathrm{Re}\left(z\right)=\frac{13769-5058\sin \theta \cos \theta }{49+9{\cos }^2\theta }\\ \end{array}$

Clearly, 13769 is divisible by 49 
 So, for $Re\left(z\right)$ to be positive integer,  $\cos \theta =0$
 Therefore, $\mathrm{Re}\left(z\right)=\frac{13769}{49}=281$