If point $R$ is placed so that $PQ = QR = PR$, then the resulting $\triangle PQR$ is equilateral. Since points $P$ and $Q$ are fixed, then there are two possible equilateral triangles with $PQ$ as a side - one on each side of $PQ$. One way to see this is to recognize that there are two possible lines through $P$ that make an angle of $60^{\circ}$ with $PQ$. Consider the line segment $PQ$. Draw a circle with centre $P$ that passes through $Q$ and a circle with centre $Q$ that passes through $P$. Suppose that the point $R$ satisfies $PQ = QR = PR$. Since $PQ = QR$, then $P$ and $R$ are the same distance from $Q$, so $R$ lies on the circle with centre $Q$ that passes through $P$. Since $PQ = PR$, then $R$ lies on the circle with centre $P$ that passes through $Q$. In other words, point $R$ is on both circles in the diagram. Since these two circles intersect in exactly two points, then there are two possible locations for $R$.