The problem can be represented as follows:

Given AB = 12 cm, AP = 3 cm, PB = 12 - 3 = 9cm.

In $\Delta PBQ, \cos \theta = \large\frac{PQ}{PB} $$ = \large\frac{PQ}{9}$

In $\Delta PRA, \sin \theta = \large\frac{PR}{PA}$$ = \large\frac{PR}{3}$

We know that $\sin^2 \theta + \cos^2\theta = 1 \rightarrow \large(\frac{PQ}{9})^2 $$ + \large(\frac{PR}{3})^2$$=1$

This gives us the equation of the locus of point P as $\large\frac{x^2}{81}$$+\large\frac{y^2}{9}$$=1$