The correct answer is $\boxed{\text{C}}$.
We can find the derivative of $f$ as follows:
\begin{align}
{\text{f’}}\left( \theta \right) &= \det \left( {\begin{array}{{20}{c}} {\cos \theta }&{-\sin \theta }&{1} \ {-\frac{1}{2}}&{\frac{\sqrt{3}}{2}}&0 \ {\frac{\sqrt{3}}{2}}&-\frac{1}{2}&1 \end{array}} \right) \\[6pt] &= \left( -\cos \theta – \frac{\sqrt{3}}{2} \right) \left( -\frac{\sqrt{3}}{2} – \frac{1}{2} \right) – \left( -\sin \theta \right) \left( \frac{\sqrt{3}}{2} \right) \\[6pt] &= \sin \theta \cos \theta + \frac{\sqrt{3}}{2} \sin \theta – \frac{\sqrt{3}}{2} \cos \theta \\[6pt] &= \frac{\sqrt{3}}{2} \left( \sin \theta – \cos \theta \right)
\end{align*}
We can see that ${\text{f’}}\left( \theta \right) = 0$ when $\sin \theta = \cos \theta$. This occurs when $\theta = \frac{\pi}{4}$ or $\theta = \frac{5\pi}{4}$. Therefore, there exists $\theta \in \left( {\frac{\pi }{6},\,\frac{\pi }{3}} \right)$ such that ${\text{f’}}\left( \theta \right) = 0$.
However, we can also see that ${\text{f’}}\left( \theta \right)$ is not continuous on $\left[ {\frac{\pi }{6},\,\frac{\pi }{3}} \right]$. This is because ${\text{f’}}\left( \frac{\pi}{4} \right)$ and ${\text{f’}}\left( \frac{5\pi}{4} \right)$ have different signs. Therefore, there exists $\theta \in \left( {\frac{\pi }{6},\,\frac{\pi }{3}} \right)$ such that ${\text{f’}}\left( \theta \right) \ne 0$.
In conclusion, both statements I and II are true.