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(C) Let the side length of the equilateral triangle be $l$. Since one vertex is at the origin $(0,0)$ and the triangle is symmetric about the $x$-axis

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$(32a, 0)$

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(C) Let the side length of the equilateral triangle be $l$. Since one vertex is at the origin $(0,0)$ and the triangle is symmetric about the $x$-axis,the other two vertices are $A\left(\frac{\sqrt{3}l}{2}, \frac{l}{2}ight)$ and $B\left(\frac{\sqrt{3}l}{2}, -\frac{l}{2}ight)$.Since vertex $A$ lies on the parabola $y^2=16ax$,we substitute its coordinates:$\left(\frac{l}{2}ight)^2 = 16a\left(\frac{\sqrt{3}l}{2}ight)$$\frac{l^2}{4} = 8\sqrt{3}al$Since $l 
eq 0$,we have $l = 32\sqrt{3}a$.Now,the coordinates of the vertices are $O(0,0)$,$A\left(\frac{\sqrt{3}(32\sqrt{3}a)}{2}, \frac{32\sqrt{3}a}{2}ight) = (48a, 16\sqrt{3}a)$,and $B(48a, -16\sqrt{3}a)$.The centroid $G$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}ight)$:$G = \left(\frac{0+48a+48a}{3}, \frac{0+16\sqrt{3}a-16\sqrt{3}a}{3}ight) = \left(\frac{96a}{3}, 0ight) = (32a, 0)$.

An equilateral triangle is inscribed in the parabola $y^2=16ax$ with one of its vertices at the origin. Then,the centroid of that triangle is

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