Let $I, I_A$ and $D$ be the incenter, $A$-excenter and the intersection of $AI$ with $BC.$ Clearly, $AKD$ is isosceles, where $K$ is the intersection of $AR$ with $BC.$ And because $K$ is the midpoint of $AR,$ we have that $\angle{ADR}=90^{\circ},$ hence $PQ$ passes through $D.$ Because $\angle{IBP}=\angle{IDP}=\angle{IDQ}=\angle{ICQ}=90^{\circ},$ we have that $I$ is the Miquel point of $BPCQI_AD,$ hence by some angle chasing, $W$ is the reflection of $I$ over $PQ$, where $W$ is the midpoint of arc $BC.$ This implies that $OI$ is perpendicular of $AW,$ which is known to be equivalent with $AB+AC=2BC,$ and $BC=4$ trivially follows.

Let $M=AR\cap BC$, $AL$ the bisector of $\triangle ABC$, $AL\cap BQ=L'$. A simple angle chasing shows that $PQ\perp AL$, so $L'A$ is a median in the right-angled triangle $ARL'$, thus $MA=ML'$. But it can be easily shown by angle chasing that $ML=AL$, and since the circle centered at $M$ with radius $MA$ intersects $AL$ in at most two points one of which is $A$, we get that $L=L'$. Let $S$ and $T$ be the midpoints of $AB$ and $AC$, respectively. Since $\angle ASQ=\angle ALQ=90^\circ$, it follows that $$ \angle BSL=\angle LQA=\angle PCA=90^\circ-\frac{\angle APC}2=90^\circ-\frac{\angle ABC}2, $$which means that the triangle $BSL$ is isosceles with $BL=SB$. Similarly, we have $CL=CT$, thus $$ 12=AB+AC+BC=2BS+2CT+BC=2BL+2CL+BC=3BC\Rightarrow BC=4. $$