What do you mean by “all $X_aX_bX_c$ triangles have a common point.”? If you mean the interior of the triangles， I think the circumcenter of $\triangle ABC$ should be the answer.

Nope! The orthocenter is the answer!

common interior point I mean, gabrupro

Why so you think the orthocenter is not an interior point? (In an acute triangle)

It's a problem for 2016-2017 year's Hungary MO， the answer is Orthocenter point.

We claim that the orthocenter $H$ of $\triangle ABC$ is always an interior point of $\triangle X_aX_bX_c$. First, we will show that $X$ is always an interior point of $\triangle X_aX_bX_c$. Let $H_a, H_b, H_c$ be the projections from $X$ to $BC, CA, AB$ respectively. Since $X$ is an interior point of $\triangle ABC$, $H_b$ and $H_c$ lie on opposite sides of line $AX$. Hence, $AH_bXH_c$ is a convex quadrilateral, so $X$ does not lie inside $\triangle AH_bH_c$. Similarly, it does not lie inside $\triangle BH_cH_a$ and $\triangle CH_aH_b$, so $X$ must be an interior point of $\triangle H_aH_bH_c$. Taking a homothety centered at $X$ with ratio $2$, we see that $X$ is always an interior point of $\triangle X_aX_bX_c$. If $X = H$, then obviously the claim is true for this case. If $X$ lies on any of the line segments $AH$, $BH$, or $CH$, WLOG let $X$ lie on line segment $AH$. Then $H$ lies on line segment $XX_a$, so $H$ is an interior point of $X_aX_bX_c$. It remains to show that the claim is true when $X$ does not lie on any of the line segments $AH$, $BH$, or $CH$. Here, $X$ is an interior point of one of the triangles $\triangle BHC, \triangle CHA, \triangle AHB$. WLOG, let $X$ be an interior point of $\triangle BHC$. Note that since $XX_b || BH$, line segments $XX_b$ and $CH$ intersect, let this intersection point be $Y_b$. Similarly, let line segments $XX_c$ and $BH$ intersect at point $Y_c$. Thus, $XY_bHY_c$ is a parallelogram. Also note that $Y_b$ and $Y_c$ are interior points of $\triangle ABC$, so $XH_b > XY_b$ and $XH_c > XY_c$. Reflect $X$ across points $Y_b$ and $Y_c$, and let the images of those reflections be $Z_b$ and $Z_c$ respectively. Since $XY_bHY_c$ is a parallelogram, $H$ lies on line segment $Z_bZ_c$. Then $XX_b = 2XH_b > 2XY_b = XZ_b$ and $XX_c = 2XH_c > 2XY_c = XZ_c$. So $Z_b$ and $Z_c$ lie on line segments $XX_b$ and $X_c$ respectively, and $H$ must be an interior point of $XX_bX_c$. And since $X$ lies inside $\triangle X_aX_bX_c$, this implies that $H$ must be an interior point of $\triangle X_aX_bX_c$, which proves the claim and completes the solution to this problem.