Since $QR$ is antiparallel to $BC$ WRT $AB,AC,$ we have $AR \cdot AC=AQ \cdot AB$ $\Longrightarrow$ $A$ has equal power WRT the circles $\odot(BQS)$ and $\odot(CRT)$ with diameters $\overline{BS}$ and $\overline{CT}$ $\Longrightarrow$ $AD \perp BC$ is radical axis of $\odot(BQS)$ and $\odot(CRT).$ Since $\angle ACB= \angle AQX=\angle XSB,$ then $XS \parallel AC.$ Similarly $YT \parallel AB,$ thus $XYTS$ is cyclic $\Longrightarrow$ $SX,TY,AD$ are then pairwise radical axes of $\odot(XYTS),$ $\odot(BQS)$ and $\odot(CRT)$ $\Longrightarrow$ $SX,TY,AD$ concur at their radical center $Z.$ Assume that $Z \in QR,$ i.e. $X \equiv Y \equiv Z.$ Then $BZ \perp ZS \parallel AC,$ $CZ \perp ZT \parallel AB,$ i.e. $Z$ is orthocenter of $\triangle ABC.$

Outline: We first show that $AD$ is the radical axis of $\odot (BQXS)$ and $\odot (CRYT).$ Afterwards, we prove that the points $S, X, Y, T$ are concyclic. It follows that $\odot (BQXS)$ and $\odot (CRYT)$ intersect twice on $AD$, and by the Radical Axis Theorem, we finish the first part. For the second part, we show that $Z \in QR \iff BZ \perp AC, \; CZ \perp AB$, which is true iff $Z = H.$ Proof. First, we claim that line $AD$ is the radical axis of $\odot (BQXS)$ and $\odot (CRYT).$ First, remark that since $\angle BQS = \angle CRT = 90^{\circ}$, it follows that $BS$ and $CT$ are diameters of $\odot (BQXS)$ and $\odot (CRYT)$, respectively. Therefore, the line joining the centers of these circles is the line $BC.$ Hence, the radical axis of these two circles is a line perpendicular to $BC.$ Therefore, if we can show that some point on the line $AD$ has equal power with respect to $\odot (BQXS)$ and $(CRYT)$, we will have shown the desired fact. We claim that point $P$ has equal power with respect to both circles. By Power of a Point, it suffices to show that $PQ \cdot PS = PR \cdot PT.$ By invoking Power of a Point again, this equality is equivalent to quadrilateral $QRST$ being cyclic. Indeed this follows from the cyclic quadrilateral $ARPQ$: We have that\[\measuredangle RTS = \measuredangle PTD = 90^{\circ} - \measuredangle DPT = 90^{\circ} - \measuredangle APR\]\[= 90^{\circ} - \measuredangle AQR = \measuredangle RQS.\] Therefore, $\measuredangle RTS = \measuredangle RQS$, so the points $Q, R, S, T$ are concyclic. Thus, we conclude that $AD$ is the radical axis of $\odot (BQXS)$ and $(CRYT).$ $\blacksquare$ Now, we claim that the points $S, X, Y, T$ are concyclic. To show this, we first prove that $SX \parallel AC.$ For these lines to be parallel, it suffices to show that $\measuredangle QRC = \measuredangle QXS.$ This follows from cyclic quadrilaterals $BQRC$ and $BQSX$, since \[\measuredangle QRC = \measuredangle QXS \iff \measuredangle QBC = \measuredangle QBS\] which is true. Thus, we have that $SX \parallel AC$, and similarly, $TY \parallel AB.$ By using these pairs of parallel lines, we have that \[\measuredangle YXS = \measuredangle QRC = \measuredangle QBC = \measuredangle YTS.\] Therefore, $\measuredangle YXS = \measuredangle YTS$, which implies that the points $S, X, Y, T$ are concyclic. $\blacksquare$ Now, since $AD$ is the radical axis of $\odot (BQSX)$ and $\odot (CRYT)$, it follows that these two circles intersect twice on the line $AD.$ Call these intersections points $U, V.$ Then, by applying the radical axis theorem to the following three pairs of concyclic points $U, V, T, Y; \quad U, V, S, X; \quad S, X, T, Y \quad$ we deduce that the lines $SX, TY, AD$ concur. This concludes the proof of the first part. $\blacksquare$ For the second part, note that since $X, Y \in QR$, the lines $SX, TY, AD$ concur on the line $QR \iff X = Y = Z.$ Let us call this triple intersection point $H'.$ We show that $H' = H$, the orthocenter of $\triangle ABC.$ Recall that the lines $SX$ and $AC$ are parallel. Therefore, since $BX \perp XS$, it follows that $BX \perp AC.$ Hence, $BH' \perp AC.$ Similarly, we find that $CH' \perp AB.$ These two relations hold iff $H'$ is the orthocenter of $\triangle ABC.$ This concludes the proof of the second part. $\blacksquare.$

My solution: Since $ QR $ is anti-parallel to $ BC $ WRT $ \angle BAC $ , so we get $ B, C, Q, R $ are concyclic and $ AB \cdot AQ=AC \cdot AR $ . ... $ (\star ) $ Since the center of $ \odot (BQS), \odot (CRT) $ is the midpoint of $ BS, CT $ , so combine with $ (\star ) $ we get $ AD $ is the radical axis of $ \{ \odot (BQS), \odot (CRT) \} $ . From Reim theorem we get $ XS \parallel AC $ and $ YT \parallel AB $ , so we get $ X, Y, T, S $ are concyclic , hence $ SX, TY, AD $ are concurrent at the radical center of $ \{ \odot (BQS), \odot (CRT), \odot (XYTS) \} $ . Since $ P $ lie on the radical axis $ AD $ of $ \{ \odot (BQS), \odot (CRT) \} $ , so we get $ PT \cdot PR=PS \cdot PQ $ . ie. $ Q, R, S, T $ are concyclic From Reim theorem we get $ BX \parallel TR $ and $ CY \parallel SQ $ , so $ Z \in QR \Longleftrightarrow X \equiv Y \equiv Z \Longleftrightarrow BZ \perp AC, CZ \perp AB \Longleftrightarrow Z \equiv H $ . Q.E.D

My solution Notice $BCPQ$ is cyclic from radical thm $A$ lie on radical axis of $(BQS)$ and $(CRT)$ $\angle XSB = \angle ACP = \angle RCS = \angle XYT \implies X,Y,T,S$ is cyclic by consider radical axis we will get $AD,XS,YT$ are concurrent at radical center $Z$. Notice $\angle YTS = \angle TXS = \angle QBS \implies AB \parallel YT$ and similary $XS \parallel AC$ if $Z$ lie on $QR$ then as $X,Y$ lie on $Q,R$ we get $X=Y=Z$.Now from $\angle BQS = \angle BZS = 90$ and so $BZ \perp AC$ similarly from other side we get $Z=H$ for $Z=H$ we get perpendicular and then it prove $Z=X=Y$