269606 At a given instant of time the position vector of a particle moving in a circle with a velocity \(3 \hat{i}-4 \hat{j}+5 \hat{k}\) is \(\hat{i}+9 \hat{j}-8 \hat{k}\). Its angular velocity at that time is:
\(\vec{\omega}=\frac{\vec{r} \times \overrightarrow{\mathrm{v}}}{r^{2}}\) \(3 \hat{i}-4 \hat{j}+5 \hat{k}=(x \hat{i}+y \hat{j}+z \hat{k}) \times(\hat{i}+9 \hat{j}-8 \hat{k})\)