Let integers $\mathrm{a}, \mathrm{b} \in[-3,3]$ be such that $\mathrm{a}+\mathrm{b} \neq 0$. 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 _____________ .

<p>$$\begin{aligned} & a, b \in I,-3 \leq a, b \leq 3, a+b \neq 0 \\ & |z-a|=|z+b| \\ & \left|\begin{array}{ccc} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{array}\right|=1 \\ & \Rightarrow\left|\begin{array}{ccc} z & z & z \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{array}\right|=1 \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow z\left|\begin{array}{ccc} 1 & 1 & 1 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{array}\right|=1 \\ & \Rightarrow z\left|\begin{array}{ccc} 1 & 0 & 0 \\ \omega & z+\omega^2-\omega & 1-\omega \\ \omega^2 & 1-\omega^2 & z+\omega-\omega^2 \end{array}\right|=1 \end{aligned}$$</p> <p>$$\begin{aligned} &\begin{aligned} & \Rightarrow z^3=1 \\ & \Rightarrow z=\omega, \omega^2, 1 \end{aligned}\\ &\text { Now }\\ &\begin{aligned} & |1-\mathrm{a}|=|1+\mathrm{b}| \\ & \Rightarrow 10 \text { pairs } \end{aligned} \end{aligned}$$</p>