In a triangle $\text{PQR, PQ} = 12$ cm and $\text{PR} = 9$ cm and $\angle \text{Q} +\angle \text{R}=120^{\circ}$. If the angle bisector of $\angle \text{P}$ meets $\text{QR}$ at $\text{M}$, find the length of $\text{PM}$
- $\dfrac{28\sqrt5}{9}$ cm
- $\dfrac{42\sqrt5}{11}$ cm
- $\dfrac{36\sqrt3}{7}$ cm
- $4\sqrt3$