Denote by $Y$ the radical center of the three circles, $PP' \cap QQ' \cap RR'$. Firstly, we will find equivalent conditions for the two conditions in the statement of the problem. Observe that $S$ lies on the polar of $PP' \cap RR'$ by Brokard, and so by La Hire, since $O_1 = \textbf{pol}(PP')$, we have that $\textbf{pol}(RR') \in SO_1$. Therefore, it's immediate that $O_3 \in SO_1$ is equivalent to $O_3 = \textbf{pol}(RR')$, i.e. $\omega_2, \omega_3$ are orthogonal. Now, observe that $\angle TQR = 90 - \angle RPQ$, and so hence $T \in QS$ is equivalent to $\angle SQR = 90 - \angle RPQ$. Since we have that: $$\angle RP'Q + \angle RPQ = \angle PRP' - \angle PQP' = 180 - \frac{\angle PO_2P'}{2} - \frac{PO_1P'}{2} = 90,$$we know that $T \in QS$ $\Leftrightarrow$ $\angle RP'Q = \angle RQS \Leftrightarrow SQ^2 = SR * SP'$, where the last part comes from similar triangles. Therefore, it suffices to show that $SQ^2 = SR * SP' = SP * SR'$ iff $\omega_2, \omega_3$ are orthogonal. Firstly, we'll prove the only if direction. Suppose that $SQ^2 = SR * SP' = SP * SR'$. Then, consider the inversion at $S$ with radius $SQ$. Clearly it fixes $\omega_2$. As $P, P', Q$ are mapped to $R', R, Q$ respectively, we have that $(\triangle PP'Q) \rightarrow (\triangle R'RQ)$, i.e., $\omega_1$ and $\omega_3$ are swapped. Since inversion preserves orthogonality and $\omega_1, \omega_2$ are orthogonal, we have that $\omega_3, \omega_2$ are orthogonal and so the only if direction is proven. Now, suppose that $\omega_2, \omega_3$ are orthogonal. Consider the circle $\omega$ centered at $S$ with radius $\sqrt{SP * SR'}$. Let it meet $\omega_2$ at $X, X'$ and $\omega_3$ at $Z, Z'$. Observe that $SX, SX'$ are tangents to $\omega_2$, and so by Brokard, since $\mathbf{pol}(S) = XX'$, we have that $Y \in XX'$. Hence, by the Radical Axis Theorem on $\omega, \omega_2, \omega_3$, we see that their radical center must be $Y$, and so inparticular $Y \in ZZ'$. Observe that by our previous work, we know that $S \in O_1O_3$, and so hence $SO_3 \perp QQ'$. Since $SO_3 \perp ZZ'$ as well, we know that $QQ' \parallel ZZ'$. Since $Y$ is on $QQ'$ by definition and it's on $ZZ'$ by the above, we must have that $QQ' = ZZ'$, i.e., $Q=Z$. This implies that $SQ^2 = SZ^2 = SR* SP'$, as desired. $\square$

We first show that $O_1,S,O_3$ are collinear if and only if $\omega_2$ and $\omega_3$ are orthogonal. Indeed, by Pascal's theorem on $PPRP'P'R'$ and $PRRP'R'R'$ in $\omega_2$ we see that $$PP\cap P'P'=O_1,\ RR\cap R'R',\ PR\cap P'R',\ PR'\cap P'R=S$$ are collinear. As $RR\cap R'R', O_2,O_3$ are collinear (on the perpendicular bisector of $RR'$), we have that $S,O_1,O_3$ are collinear if and only if either $O_3=RR\cap R'R'$ or $O_1,O_2,O_3$ are collinear. If $O_1,O_2,O_3$ are collinear our clockwise orientation cannot hold (we have that diametrically opposite vertices of an isosceles trapezoid must be consecutive), so $O_1,S,O_3$ are collinear if and only if $O_3=RR\cap R'R'$. This is equivalent to $O_3R\perp RO_2$ and $O_3R'\perp R'O_2$, which is equivalent to $\omega_2$ and $\omega_3$ being orthogonal. Now, we show that $\omega_3\perp \omega_2$ if and only if $T,Q,S$ are collinear. We first prove the following lemma: Lemma. In triangle $ABC$, a circle $\gamma$ passes through $B$ and $C$. An inversion is performed with center $A$ that sends $B$ to $B_1$ and $C$ to $C_1$. The image of $\gamma$ is the circle with diameter $B_1C_1$ if and only if $\gamma$ and $(ABC)$ are orthogonal. Proof of LemmaProof. Let $AB$ and $AC$ intersect $\gamma$ again at $F$ and $E$, respectively. Now, perform the inversion with radius $\sqrt{\mathrm{Pow}_{\gamma}(A)}$ so that $B\to F$ and $C\to E$, so that $\gamma$ is fixed. Now, if $EF$ is a diameter of $\gamma$, $EB$ and $FC$ are altitudes in $\triangle AEF$. The center of $(ABC)$ is thus the midpoint $N$ of $AH$ ($H$ is the orthocenter of $AEF$) and the center of $\gamma$ is the midpoint $M$ of $EF$, and $MB\perp NB$ and $MC\perp NC$ as $MN$ is a diameter of the nine-point circle of $AEF$, so $\gamma$ and $(ABC)$ are orthogonal. On the other hand, if $\gamma$ and $(ABC)$ are orthogonal then $O$, the center of $\gamma$, is the intersection of the tangents at $B$ and $C$ to $(ABC)$. We have \begin{align*} \angle EOF &= \angle EOC+\angle BOF\\ &=\left(180^{\circ}-2\angle ECO\right)+\left(180^{\circ}-2\angle FBO\right)+\angle COB\\ &=\left(2\angle ABO - 180^{\circ}\right)+\left(2\angle ACO - 180^{\circ}\right)+\angle COB\\ &=2\angle B + 2\angle C + 2\left(\angle BCO+\angle CBO\right) - 360^{\circ}+\angle COB\\ &=2\angle B + 2\angle C - \angle COB\\ &=360^{\circ}-2\angle A - \left(180^{\circ}-2\angle A\right)\\ &=180^{\circ}, \end{align*} so $EF$ is a diameter of $\gamma$, finishing the proof. Now, perform an inversion about $Q$, and define points so that $P\to A$, $R\to B$, $P'\to C$, $R'\to D$, $Q\to X$, $S\to Y$. As $(QPP')=\omega_1\perp \omega_2,$ we have by our lemma that $\omega_2$ inverts to the circle with diameter $AC$. We have that $T,Q,S$ are collinear if and only if $QS$ and the line through $Q$ perpendicular to $PR$ are isogonal in $\angle PQR$; under inversion this is equivalent to $QY\perp AB$. As $S=P'R\cap PR'$, $Y=(BCQ)\cap (ADQ)$; applying radical center to these two circles and $(ABCD)$ gives that $QY$, $AD$, and $BC$ concur. As $B$ and $D$ are on the circle with diameter $AC$, we have that $BC\perp AB$, so $QY\perp AB$ happens if and only if the three lines concur at the infinity point along $BC$, or equivalently $AD||BC$. This is equivalent to $AD\perp AB$, or $BD$ being a diameter of $(ABCD)$, which by our lemma happens only if $(QRR')=\omega_3$ and $\omega_2$ are orthogonal, finishing the proof.