The problem is related to the post Prove: A triangle inscribed in a rectangular hyperbola has its orthocenter on that hyperbola, the post Prove that a conic section through the vertices of a triangle and its orthocentre is a rectangular hyperbola., Wolfram MathWorld entry Circumhyperbola and Wikipedia entry Nine-point hyperbola. From the last sentence of an image:
For any hyperbola passing through the orthocenter of a triangle and the circumcircle, the line connecting the antipode of the fourth intersection point with the centroid passes through the center of the hyperbola.
I attempt to give a proof. Working in homogeneous barycentric coordinates $(x\!:\!y\!:\!z)$ with respect to $\triangle ABC$,
$$ A=(1\!:\!0\!:\!0),\quad B=(0\!:\!1\!:\!0),\quad C=(0\!:\!0\!:\!1), $$
the circumcircle is
$$ a^2\,yz \;+\;b^2\,zx\;+\;c^2\,xy\;=\;0,\tag{1} $$
and the centroid is
$$ G=(1\!:\!1\!:\!1)\,. $$
The orthocenter is
$$ H=(\tan A:\tan B:\tan C)\,, $$
Recall that the centre of a conic is pole to the line at infinity, the line at infinity is
$$ x+y+z=0\,. $$
Hence if a circumconic has matrix $M$, its centre $K$ is the solution of
$$ M\,(1,1,1)^{T}\;=\;0\,. $$
From https://www.maths.gla.ac.uk/wws/cabripages/triangle/newconic2.htm,
If a non-degenerate circumconic is perspective from a point $P=(p:q:r)$, then its centre is
$$K \;=\;\bigl(p\,(q+r-p)\;:\;q\,(r+p-q)\;:\;r\,(p+q-r)\bigr)\,. $$
Here “perspective from $P$” means that the cevians $AP,BP,CP$ are the polars of the contact‐triangle of the conic. In particular every rectangular circumhyperbola through $H$ arises this way, with some $P$ lying on the circumcircle (and conversely).
A rectangular circumhyperbola meets the circumcircle in the three vertices $A,B,C$ and in one further point $X$, point $X$ is in fact the perspector of $\mathcal H$
Hence if we write
$$ X=(x:y:z)\quad\bigl(a^2yz+b^2zx+c^2xy=0\bigr), $$
then the centre of the unique rectangular circumhyperbola through $A,B,C,H$ and $X$ is
$$ K \;=\;\bigl(x\,(y+z-x)\;:\;y\,(z+x-y)\;:\;z\,(x+y-z)\bigr)\,. $$
Let $X' =$ the antipode of $X$ on the circumcircle. In vector (or trilinear) language one checks easily that
$$ X' \;=\;(y+z-x\;:\;z+x-y\;:\;x+y-z)\,, $$
and that the centroid is $G=(1:1:1)$. So
$$ G \;=\;(1:1:1),\quad K \;=\;\bigl(x\,(y+z-x)\;:\;y\,(z+x-y)\;:\;z\,(x+y-z)\bigr),\quad X'=\bigl(y+z-x\;:\;z+x-y\;:\;x+y-z\bigr). $$
But now the three vectors
$$ (1,1,1),\quad (x\alpha,y\beta,z\gamma),\quad(\alpha,\beta,\gamma), $$
with $\alpha=y+z-x$, $\beta=z+x-y$, $\gamma=x+y-z$, are manifestly coplanar in $\Bbb R^3$, hence their projective representatives $G,K,X'$ are collinear. Equivalently,
$$ \det \begin{pmatrix} 1&1&1\\[6pt] x\alpha & y\beta & z\gamma\\[3pt] \alpha & \beta & \gamma \end{pmatrix} \;=\;0, $$
