Angular momentum of the particle about origin O will be given by



$\overrightarrow{\text{L}}=\overrightarrow{\text{r}}\times \overrightarrow{\text{p}}=\text{m}\left(\overrightarrow{\text{r}}\times \overrightarrow{\text{V}}\right)$

$\text{or}\left|\overrightarrow{\text{L}}\right|=\text{L}=\text{mrV sin}\mathrm{\theta }$

$=\text{mV}\left(\text{r sin}\mathrm{\theta }\right)=\text{mVh}$

Now, $\text{m,V}$and $\text{h}$all are constant

Therefore, angular momentum of particle about origin will remain constant. The direction of $\overrightarrow{\text{r}}\times \overrightarrow{\text{v}}$ also remains the same (negative z direction)