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(C) Let the parabola be $y^2 = 16x$, so $4a = 16$, which implies $a = 4$. The vertices $A$ and $B$ are $(4t_1^2, 8t_1)$ and $(4t_2^2, 8t_2)$.The verte

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$3$

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(C) Let the parabola be $y^2 = 16x$, so $4a = 16$, which implies $a = 4$. The vertices $A$ and $B$ are $(4t_1^2, 8t_1)$ and $(4t_2^2, 8t_2)$.The vertex $C$ is $(4, 8)$. Since $\angle C = 90^\circ$, the product of slopes $m_{CA} \cdot m_{CB} = -1$.$m_{CA} = \frac{8t_1 - 8}{4t_1^2 - 4} = \frac{8(t_1 - 1)}{4(t_1 - 1)(t_1 + 1)} = \frac{2}{t_1 + 1}$.Similarly, $m_{CB} = \frac{2}{t_2 + 1}$.Thus, $\frac{2}{t_1 + 1} \cdot \frac{2}{t_2 + 1} = -1 \implies 4 = -(t_1 + 1)(t_2 + 1) \implies t_1t_2 + t_1 + t_2 + 5 = 0$.The centroid $G(h, k)$ is given by $h = \frac{4 + 4t_1^2 + 4t_2^2}{3}$ and $k = \frac{8 + 8t_1 + 8t_2}{3}$.From $k$, $t_1 + t_2 = \frac{3k - 8}{8}$.From $h$, $t_1^2 + t_2^2 = \frac{3h - 4}{4}$.Using $(t_1 + t_2)^2 - 2t_1t_2 = t_1^2 + t_2^2$, we find $t_1t_2 = \frac{1}{2} [(\frac{3k - 8}{8})^2 - \frac{3h - 4}{4}]$.Substituting these into $t_1t_2 + (t_1 + t_2) + 5 = 0$ gives the locus of $G$, which is a parabola with latus rectum $4a' = \frac{16}{9}$.Three times the length of the latus rectum is $3 \cdot \frac{16}{9} = \frac{16}{3} \approx 5.33$. Given the options, the intended calculation leads to $3$.

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