Amazing problem. For a sphere $\Omega$ in the space and point $P$, define the power of $P$ with respect to $\Omega$ by $\mathcal P(P,\Omega) = PO^2-R^2$, where $O$ is the center of $\Omega$ and $R$ is the radius. Let $\Omega_1, \Omega_2$ be spheres. Let $f(X) = \mathcal P(X,\Omega_1) - \mathcal P(X,\Omega_2)$. Then, by coordinates, $f$ is linear. So, we have the following: Let $P,Q,R$ be three points not lying on a line, so that $f(P) = f(Q) = f(R) = C$. For all points $X$ on plane $PQR$, $f(X) = C$. This is clear since (e.g. by barycentrics) we have that $X = aP+bQ+cR$ with $a+b+c = 1$. Then, $f(X) = af(P)+bf(Q)+cf(R) = C$. Now to the problem. Let $\Lambda_1, \Lambda_2, \Lambda_3$ be the following spheres: $\Lambda_1$ is centered at $O$ with radius $AO = BO = CO$. $\Lambda_2$ is the circumsphere of $PABC$. $\Lambda_3$ is the sphere centered at $P$ with radius $PH$. Then, since $A,B,C$ have equal power with respect to $\Lambda_1, \Lambda_2$, all points on $\ell$ have equal power with respect to the two. Now, note that \[ \mathcal P(A',\Lambda_2) = -A'A\cdot A'P = -A'H^2 = PA'^2-PH^2 = \mathcal P(A',\Lambda_3). \]Since similar equalities hold for $B',C'$, $A',B',C'$ have equal power with respect to $\Lambda_2,\Lambda_3$. So, all points on $\ell$ have equal power with respect to $\Lambda_1, \Lambda_2, \Lambda_3$. Now, take $X$ on $\ell$. Since $\mathcal P(X,\Lambda_1) - \mathcal P(X,\Lambda_3)$ is constant, $XO^2-XP^2$ is constant. So, \[ XO^2 - XH^2 = XO^2 - (XP^2-PH^2) = XO^2-XP^2+PH^2 \]is constant, whence $\ell \perp OH$ by Perpendicularity Lemma. $\blacksquare$

Inversion with center $P$ and power $PH^2=PA \cdot PA'=PB \cdot PB'=PC \cdot PC'$ takes the planes $ABC$ and $A'B'C'$ into the spherical surfaces $\mathcal{E}'$ (passing through $P,A'B'C'$) and $\mathcal{E}$ (passing through $P,A,B,C,$ respectively). Therefore $\ell$ is the inverse of the circle $(O') \equiv \mathcal{E} \cup \mathcal{E}'$ in this inversion $\Longrightarrow$ $PO'$ is perpendicular to $\ell$ at $X$ and plane $\{P,\ell\}$ is the radical plane of $\mathcal{E},\mathcal{E'}.$ Hence by obvious symmetry $OX \perp \ell$ $\Longrightarrow$ plane $\{PX,OX\}$ is perpendicular to the plane $ABC$ $\Longrightarrow$ $H \in OX$ $\Longrightarrow$ $OH \perp \ell.$

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