[asy][asy] size(8cm); pair A = dir(105); pair B = dir(210); pair C = dir(330); pair R = foot(A,B,C); path w = CP(A,R); pair P = intersectionpoints(w,unitcircle)[0]; pair Q = intersectionpoints(w,unitcircle)[1]; pair K = -A; draw(K--(0,0),dashed); draw(P--K--Q); draw(A--B--C--A--cycle); draw(unitcircle); draw(P--Q); draw(A--(0,0)); draw(A--R); draw(w); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$O$",(0,0),dir(45)); dot("$K$",K,dir(K)); [/asy][/asy] The main claim is that $P$ and $Q$ are precisely the intersection points of the circles $\omega$ and $\Gamma$. To show this, we invert at $R$. Let $X^*$ denote the inverse of point $X$. Since $AB<AC$, this means that $A^*B^*>A^*C^*$. [asy][asy] size(8cm); pair R = (0,0); pair B = (-2,0); pair C = (1.5,0); pair A = (0,2.4); pair P = (-2.5,1.2); pair Q = (2,1.2); pair X = intersectionpoints(P--Q,circumcircle(A,B,C))[0]; pair Y = intersectionpoints(P--Q,circumcircle(A,B,C))[1]; draw(circumcircle(A,B,C)); draw(A--P--Q--A--cycle); draw(B--C); draw(A--R); draw(P--C); draw(B--Q); dot("$A^*$",A,dir(A)); dot("$B^*$",B,dir(B)); dot("$C^*$",C,dir(C)); dot("$R$",R,dir(270)); dot("$P^*$",P,dir(P)); dot("$Q^*$",Q,dir(Q)); dot("$X$",X,dir(45)); dot("$Y$",Y,dir(135)); [/asy][/asy] We first translate the two conditions in the question. First, for $\measuredangle BQR=\measuredangle RPC$, we obtain \[\measuredangle Q^*B^*R=\measuredangle RC^*P^*.\]As for the second condition, $AO\perp PQ$, we refer to our original diagram to see that since $AP=AQ$, this means that we can also write this condition as $\measuredangle PAO = \measuredangle OAQ$. Therefore, \[\measuredangle PAR + \measuredangle RAO = \measuredangle OAR + \measuredangle RAQ.\]Since $\measuredangle RAO = \measuredangle BAO- \measuredangle BAR = (90^\circ - \measuredangle ACB) - (90^\circ - CBA) = \measuredangle CBA - \measuredangle ACB$, we obtain \[\measuredangle PAR + \measuredangle QAR + 2(\measuredangle RBA - \measuredangle ACR) = 0.\]Translating into the inverted diagram, \[\measuredangle A^*P^*R + \measuredangle A^*Q^*R + 2(\measuredangle RA^*B^* - \measuredangle C^*A^*R) = 0.\]Since the circle $\omega$ is tangent to $BC$ at $R$, this means that $\omega^*$ is parallel to $B^*C^*$, and in fact it is the perpendicular bisector of $A^*R$ (to see this, consider the point diametrically opposite $R$ on $\omega$, it is mapped to the midpoint of $A^*R$). Therefore, $P^*Q^*C^*B^*$ is an isosceles trapezoid, and $R$ is the reflection of $A^*$ across $P^*Q^*$. The angle condition reads \[2(90^\circ-\measuredangle RA^*P^*+90^\circ - \measuredangle RA^*Q^* + \measuredangle RA^*B^* + \measuredangle RA^*C^*=0\]Therefore, \[2(\measuredangle B^*A^*P^*+\measuredangle C^*A^*Q^*)=0.\]If we let $X$ and $Y$ be the intersection points of $P^*Q^*$ with the circumcircle of $\triangle A^*B^*C^*$, then we know that $\angle XA^*B^* = \angle C^*A^*Y$, therefore \[\measuredangle P^*A^*X+\measuredangle YA^*Q^*=0\]Suppose each of the angles is nonzero. Since $P^*X=Q^*Y$ by symmetry about the perpendicular bisector of $B^*C^*$, this can only mean that $A^*P^*=A^*Q^*$ or that $A^*$ lies on the perpendicular bisector of $B^*C^*$, contradicting the assumption that $A^*B^*\neq A^*C^*$. Therefore, both angles are equal to zero, and $P^*$ and $Q^*$ lie on $\omega^*$. Therefore, $P$ and $Q$ lie on $\omega$. For the finish, note that if $K$ is the point diametrically opposite $A$ on $\omega$, then $\angle APK=\angle AQK=90^\circ$, therefore $PK$ and $QK$ are tangents to $\Gamma$.