R is the circumcenter, H is the orthocenter of the triangle ABC.
Then points F,G,E are the points of intersection of the altitudes with the sides of the triangle ABC. U, V, W are the intersection points of cevians AR, BR, CR with the sides of the triangle ABC.
X is defined as the intersection of lines GU, EF. Y is defined as the intersection of GU, WV and Z is the intersection point of lines WV, EF. Then it can be shown and proven that lines AX, BY, CZ always intersect at the point RH.
UPDATE: Actually, 4 'RH' points exist. **3 of them always belong to Jerabek hyperbola that goes through A,B,C,R,H and the fourth RH point lies on that hyperbola only in a special case. All the points of this configuration are shown here.
I wonder what is known about these RH points, K points, S point and related conics?
In fact, 3 RH points can be any of those or completely new ones:
54 (Kosnita point), 64 isogonal conjugate of the de Longchamps point), 65 (orthocenter of the contact triangle), 66 (isogonal conjugate of the Exeter point), 67 (isogonal conjugate of the far-out point), 68 (Prasolov point), 69, 70, 71, 72, 73, 74, 248, 265, 290, 695, 879, 895, 1173, 1175, 1176, 1177, 1242, 1243, 1244, 1245, 1246, 1439, 1798, 1903, 1942, 1987, 2213, 2435, 2574, 2575, 2992, and 2993.
Referring back to my previous answer to a related question, we'll consider things in a bit more generality, and handle an ambiguity in the formulation of this property.
Given $\triangle ABC$ we define points $P_+$ and $P_-$ (which correspond to the orthocenter and circumcenter in OP's current question) via barycentric coordinates $$P_\pm = \frac{\alpha_\pm A+\beta_\pm B+\gamma_\pm C}{\alpha_\pm+\beta_\pm+\gamma_\pm} \tag{1}$$ Cevians through $P_+$ and $P_-$ meet the sides of the triangle in six points, $D_\pm$, $E_\pm$, $F_\pm$ (OP's $F$, $G$, $E$, $U$, $V$, $W$), and pairs of these points determine lines that meet pairwise at three more points, $D$, $E$, $F$ (OP's $X$, $Y$, $Z$). Finally, although there's ambiguity in choosing pairs of points that ultimately determine $D$, $E$, $F$, the cevians $\overleftrightarrow{AD}$, $\overleftrightarrow{BE}$, $\overleftrightarrow{CF}$ always concur. As per my other answer, the points of concurrency (which I denote with sub-scripted (or not) $K$) have barycentric coordinates $$\begin{align} K_A &= \left(\frac12:\frac{1}{\dfrac{\alpha_+}{\beta_+}+\dfrac{\alpha_-}{\beta_-}}:\frac{1}{\dfrac{\alpha_+}{\gamma_+}+\dfrac{\alpha_-}{\gamma_-}}\right) \tag2\\[4pt] K_B &= \left(\frac{1}{\dfrac{\beta_+}{\alpha_+}+\dfrac{\beta_-}{\alpha_-}}:\frac12:\frac{1}{\dfrac{\beta_+}{\gamma_+}+\dfrac{\beta_-}{\gamma_-}}\right) \tag3 \\[4pt] K_C &= \left(\frac{1}{\dfrac{\gamma_+}{\alpha_+}+\dfrac{\gamma_-}{\alpha_-}}:\frac{1}{\dfrac{\gamma_+}{\beta_+}+\dfrac{\gamma_-}{\beta_-}}:\frac12\right) \tag4 \\[4pt] K\phantom{_X} &= \left(\frac1{\beta_+\gamma_- + \beta_-\gamma_+}: \frac1{\gamma_+\alpha_-+\gamma_-\alpha_+}:\frac1{\alpha_+\beta_-+\alpha_- \beta_+}\right) \tag5 \end{align}$$ It's not clear which of these points OP intends by "$RH$", but $75\%$ of the time, it doesn't matter. :) To see why, note that the five-point conic through $A$, $B$, $C$, $P_+$, $P_-$ has barycentric equation
$$\frac{\alpha_+}{x} \left(\frac{\beta_+}{\beta_-} - \frac{\gamma_+}{\gamma_-}\right) + \frac{\beta_+}{y}\left( \frac{\gamma_+}{\gamma_-} - \frac{\alpha_+}{\alpha_-}\right) +\frac{\gamma_+}{z}\left( \frac{\alpha_+}{\alpha_-} -\frac{\beta_+}{\beta_-}\right)= 0 \tag{6}$$ where $x:y:z$ are the barycentric coordinates of any point on the conic. The reader can verify that $K_A$, $K_B$, $K_C$ (one of which is what I suspect OP intends to be point "$RH$") automatically satisfy $(6)$, and that $K$ satisfies it only if $$\left( \frac{\alpha_+}{\alpha_-}-\frac{\beta_+}{\beta_-} \right)\left( \frac{\beta_+}{\beta_-}-\frac{\gamma_+}{\gamma_-}\right)\left(\frac{\gamma_+}{\gamma_-}-\frac{\alpha_+}{\alpha_-}\right)=0 \tag{7}$$
(In a comment to the original question, I included squares in a formula corresponding to the above. My second pass at the problem seems to make the squares unnecessary. I'm going with that.)
Observe that if, say, the first factor of $(7)$ vanishes, equivalently, if $\alpha_+/\beta_+=\alpha_-/\beta_-$ then (via my previous answer) cevian points $F_+$ and $F_-$ coincide. Consequently we can interpret $(7)$ in general as the condition that $\overleftrightarrow{P_+P_-}$ passes through a vertex of $\triangle ABC$.
In OP's specific case using the circumcenter and orthocenter, $$\begin{align} R &= (\alpha_+:\beta_+:\gamma_+) = (\sin 2 A: \sin 2 B : \sin 2C ) \\ H &= (\alpha_-:\beta_-:\gamma_-) = (\tan A:\tan B:\tan C) \end{align} \tag8$$ condition $(7)$ transforms to $$(\cos^2A-\cos^2B)(\cos^2B-\cos^2C)(\cos^2C-\cos^2A)=0 \tag{9}$$ so that (non-degenerate) $\triangle ABC$ must be isosceles for the point $K$ to lie on the conic.
Regarding the "triangle center-ness" of these $K$-points, we can note that, even in general, $K_A$, $K_B$, $K_C$ are typically not triangle centers in the Kimberling sense, as they aren't symmetrically-defined by the parent triangle. As formulas $(2)$, $(3)$, $(4)$ show, the point you get depends upon how the vertices are labeled. (That's why the subscripts make sense!)
On the other hand, point $K$ (via $(5)$) is symmetrically-defined, and it therefore generally corresponds to a triangle center. In fact, it turns out that this point is known as the Ceva Point for points $P_+$ and $P_-$. In particular, the Ceva Point for the orthocenter and circumcenter is designated $X(1105)$ in Kimberling's Encyclopedia of Triangle Centers.
The point where $\overleftrightarrow{AK_A}$, $\overleftrightarrow{BK_B}$, $\overleftrightarrow{CK_C}$ concur —which I call $K_\star$ in my previous answer— is also symmetrically-defined, so it counts as a "triangle center" of some kind. In fact, it is known as the crosspoint of $P_+$ and $P_-$. Its orthocenter-circumcenter incarnation is Kimberling's $X(185)$. (Interestingly, the ETC notes that this point is the orthocenter of the tangential triangle of the Jerabek hyperbola.)