Define $K$ to be the point on $YZ$ such that $HK \perp YZ$. Let $A'$ be a point on the circumcircle of $\triangle ABC$ such that $AA' \parallel BC$.

Lemma 1. $YZ$ is the perpendicular bisector of $HA'$.
Proof. Let $H_B$ denote the reflection of $H$ in $AC$. Then note that $A'H_B \parallel CO$. This is because $\angle (AA', A'H_B) = \angle 90^\circ - A = \angle (CO, BC)$. Now $YO \perp CO$ implies $OY \perp A'H_B$. This shows $YA' = YH_B = YH$. Similarly $ZH = ZA'$. $\square$

Lemma 2. $OK \perp BC$.
Proof. Let $D$ and $D'$ denote the foot of perpendiculars from $H, A'$ onto $BC$ respectively. Then note that if $M$ is the foot of perpendicular from $K$ onto $BC$, then by similarity, $DM = MD'$ as $HK = KA'$. However $BD = D'C$, so $MB = MC$, as desired. $\square$

Lemma 3. $RY$ and $QZ$ are tangents to the circumcircle of $\triangle AYZ$.
Proof. Note that $RKYO$ is a cycle quadrilateral. Now we angle chase: $\angle RYK = \angle ROK = \angle (CO, OK) = \angle BAC = \angle YAZ$. Thus, $RY$ is tangent to the circumcircle of $\triangle AYZ$. Similarly for $QZ$. $\square$

Finally note that $O$ is the Miquel point of quadrilateral $QZYR$, as $RKYO$ and $QKZO$ are cyclic. Thus $P := QZ \cap YR$ lies on the circumcircle of $\triangle QOR$, as desired. $\square$

Observe that $\angle BOZ + \angle COY = 180^\circ$ hence the isogonal conjugate $Q$ of point $O$ in quadrilateral $BCYZ$ exists. Since $H$ and $O$ are isogonal conjugates in $ABC$, $Q=H$ follows.
In particular, $O$ is the centre of the circle passing through the reflections of $H$ in $BC, CY, YZ, ZB$; which coincides with the circumcircle of $ABC$. Thus reflection $P$ of $H$ in $YZ$ lies on $(ABC)$.
However, $\angle YHZ = \angle A$ as $\angle YOZ = 180^\circ - 2\angle A$ and $H, O$ are isogonal conjugates in $BCYZ$, hence $P$ lies on $(AYZ)$ as well.
Now $BZ = OB/\sin C$ hence $BZ \cdot BA = 2R^2 = CY \cdot CA$, where $R$ is the circumradius of $ABC$. Thus, $B$ and $C$ have equal powers in $(AYZ)$ or the circumcentre $O_1$ of triangle $AYZ$ lies on the perpendicular bisector of $BC$, which passes through $O$.
Further, $O$ is the centre of spiral similarity sending $YZ$ to $QR$ so it suffices to show that it sends $O_1$ to $T$, where $T$ is the point of intersection of $Y$ and $Z$ tangents to $AYZ$.
Let $OO_1 \cap YZ = K$ and the line through $O$ parallel to $BC$ meet $QR$ at $S$. Now the same spiral similarity maps $K$ to $S$ so we just need to show that $OS/OK = OT/OO_1$. Both are equal to $\tan \angle (YZ, BC)$, so we are done.
To conclude, suffices to show $OD \perp YZ$ where $D$ is the foot of $A$-altitude which will allow $O, K, YZ \cap QR$ to be collinear. This follows as $HP \perp YZ$ as $H, O$ isogonal conjugates in $BCYZ$ hence
$$\angle PAY = \angle PZY = \angle HZY = \angle OZB = \angle C.$$
$\square$

We prove Lemma 2 of Solution A using complex numbers, and then finish the same way as in Solution A.
We toss the diagram on the complex plane. Let the circumcircle of $\triangle ABC$ be the unit circle, so $O$ is the origin, and let $A = a$, $B = b$, $C = 1/b$ without loss of generality. Use $z - a = -b(z\bar{x} - 1)$ and $z = -b^2\bar{z}$ to get that $Z = z = \frac{b(a+b)}{b-a}$. By symmetry, $Y = y = \frac{c(a+c)}{c-a} = \frac{(ab+1)}{b(1-ab)}$.

Now, redefine $K$ to be intersection of the perpendicular from $O$ to $BC$ with $YZ$. We need to show that $HK \perp YZ$.
We calculate
$$y - z = \frac{(1 - b^2)(ab(a + b) + a - b)}{b(1 - ab)(a - b)}, \quad \text{so} \quad \frac{y - z}{\bar{y} - \bar{z}} = \frac{ab(a + b) + a - b}{a + b + ab(b - a)}$$
Therefore, $k = z - \frac{ab(a + b) + a - b}{a + b + ab(b - a)}(\bar{k} - \bar{z})$. Also $k = \bar{k}$ since it lies on the real axis. Therefore,
$$\frac{2a + 2ab^2}{a + b + ab(b - a)} \cdot k = z \left(1 - \frac{ab(a + b) + a - b}{b^2(a + b + ab(b - a))}\right) \Rightarrow k = \frac{b(a + b)}{b - a} \cdot \frac{(b^2 + 1)(1 + ab)(b - a)}{2ab^2(1 + b^2)} = \frac{(a + b)(ab + 1)}{2ab}$$
Finally, the orthocenter $H = h = a + b + \frac{1}{b}$. Thus, $h - k = \frac{ab(a + b) + (a - b)}{2ab}$. Finally,