Construct a triangle $PQR$ given that $QR = 3 \, cm$,$\angle PQR = 45^{\circ}$ and $QP - PR = 2 \, cm$. Provide the steps of construction and justification.

(N/A) Steps of Construction:
$1.$ Draw a ray $QX$ and cut off a line segment $QR = 3 \, cm$.
$2.$ At $Q$,construct $\angle YQR = 45^{\circ}$.
$3.$ From the ray $QY$,cut off a line segment $QS = 2 \, cm$.
$4.$ Join $RS$.
$5.$ Draw the perpendicular bisector of $RS$. Let it intersect $QY$ at point $P$.
$6.$ Join $PR$. Then,$\triangle PQR$ is the required triangle.
Justification:
Since $P$ lies on the perpendicular bisector of $RS$,$PS = PR$.
Now,$QS = QP - PS = QP - PR$.
Since $QS = 2 \, cm$,we have $QP - PR = 2 \, cm$. This justifies the construction.