Construct a triangle $PQR$,in which $QR = 9\, cm$,$\angle Q = 60^{\circ}$ and $PR - PQ = 3\, cm$.

(N/A) Steps of construction:
$(1)$ Draw a ray $QX$ and cut off a line segment $QR = 9\, cm$ from it.
$(2)$ At point $Q$,construct a ray $QY$ such that $\angle YQR = 60^{\circ}$.
$(3)$ Produce the ray $QY$ backwards to form a ray $QZ$. Cut off a line segment $QS = 3\, cm$ from $QZ$.
$(4)$ Join $RS$. Draw the perpendicular bisector of the line segment $RS$.
$(5)$ Let the perpendicular bisector intersect the ray $QY$ at point $P$. Join $PR$.
$(6)$ Thus,$\Delta PQR$ is the required triangle.