Elegantly latticed triangle centers

Question

Take a non-right scalene triangle and calculate triangle centers. How many distinct lattice points in the vicinity of the origin can be covered by these triangle centers?

The triangle $ ((4\sqrt3, 3\sqrt3), (-4\sqrt3, -3\sqrt3), (3, 12))$ is the best I've found so far, back-solving from the incenter, orthocenter, and centroid. Green points are rational centers, red points are non-rational centers.

Coordinates of the first 13 centers are $$((3,6), (1,4), (-3,4), (9,4), (3,4), (89,148)/21,(183,264)/43)$$ $$( (-3,0), (-27,126)/43, (0,3), (3,9), (12,21)/4, (77,100)/17)$$

Are there any triangles with centers that will cover more distinct lattice points near the origin than this triangle?

Answer 1

Here are $3$ candidates (coordinates of vertices and draft sketches), but only $112$ first centers have been considered/drawn here yet.

First and second triangles are similar ($\sqrt{2}$-scaled and $45^\circ$-rotated).

First triangle: $(14,\;-2), \quad (-7-3\sqrt{7},\; 1+3\sqrt{7}), \quad (-7+3\sqrt{7},\; 1-3\sqrt{7})$;

Few first integer centers: $X_1(2,-2)$, $X_2(0,0)$, $X_3(2,10)$, $X_4(-4,-20)$, $X_5(-1,-5)$, $X_8(-4,4)$, $X_{10}(-1,1)$, $X_{11}(5,1)$, $X_{12}(1,-3)$.

Second triangle: $(-3,\; 3\sqrt{7}+4), \quad(-3, \;4-3\sqrt{7}),\quad (6,\;-8)$;

Few first integer centers: $X_1(0,-2)$, $X_2(0,0)$, $X_3(6,4)$, $X_4(-12,-8)$, $X_5(-3,-2)$, $X_8(0,4)$, $X_{10}(0,1)$, $X_{11}(3,-2)$, $X_{12}(-1,-2)$.

Third triangle: $(x_A,\; y_A), \quad(x_B, \;y_B),\quad (x_C,\;y_C)$,
where
$x_A,x_B,x_C$ are roots of cubic equation $x^3-39x-16=0$;
$y_A,y_B,y_C$ are roots of cubic equation $x^3-147x+578=0$;
$(x_A,y_A) \approx(-0.412050259613482,\; 9.15947187400213)$;
$(x_B,y_B) \approx(-6.02876924833151,\; 4.58967278123278)$;
$(x_C,y_C) \approx( 6.44081950794499,\; -13.7491446552349)$;

Few first integer centers: $X_1(-2,4)$, $X_2(0,0)$, $X_3(4,-2)$, $X_4(-8,4)$, $X_5(-2,1)$, $X_8(4,-8)$, $X_{10}(1,-2)$, $X_{11}(-2,7)$, $X_{12}(-2,3)$.

(Free for further research and improve).