Let integers $\mathrm{a}, \mathrm{b} \in[-3,3]$ be such that $\mathrm{a}+\mathrm{b} eq 0$. Then the number of all possible ordered pairs (a, b), for which $\left|\frac{z-\mathrm{a}}{z+\mathrm{b}}\right|=1$ and $\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1, z \in \mathrm{C}$, where $\omega$ and $\omega^2$ are the roots of $x^2+x+1=0$, is equal to $\underline{\hspace{2cm}}$ .