Let A,B and C be the vertices of a variable right angled triangle...

The equation of the parabola is $y^2 = 16x$, which gives $a = 4$. Any point on the parabola can be taken as $(4t^2, 8t)$.

The vertex $B$ is given as $(4,8)$, which corresponds to the parameter $t = 1$. Let the parameters for vertices $A$ and $C$ be $t_1$ and $t_2$ respectively. Thus, $A \equiv (4t_1^2, 8t_1)$ and $C \equiv (4t_2^2, 8t_2)$.

The slope of $AB$ is:
$m_{AB} = \dfrac{8t_1 - 8}{4t_1^2 - 4} = \dfrac{2(t_1 - 1)}{(t_1 - 1)(t_1 + 1)} = \dfrac{2}{t_1 + 1}$

Similarly, the slope of $BC$ is:
$m_{BC} = \dfrac{2}{t_2 + 1}$

Since $\triangle ABC$ is right-angled at $B$, $AB \perp BC$. Therefore, $m_{AB} \times m_{BC} = -1$:
$\dfrac{2}{t_1 + 1} \times \dfrac{2}{t_2 + 1} = -1$
$(t_1 + 1)(t_2 + 1) = -4$
$t_1t_2 + t_1 + t_2 + 5 = 0$

Let the centroid of $\triangle ABC$ be $(h, k)$. Then:
$h = \dfrac{4 + 4t_1^2 + 4t_2^2}{3} \Rightarrow t_1^2 + t_2^2 = \dfrac{3h}{4} - 1$
$k = \dfrac{8 + 8t_1 + 8t_2}{3} \Rightarrow t_1 + t_2 = \dfrac{3k}{8} - 1$

From the perpendicularity condition, we can express $t_1t_2$ as:
$t_1t_2 = -5 - (t_1 + t_2) = -5 - \left(\dfrac{3k}{8} - 1\right) = -4 - \dfrac{3k}{8}$

Using the algebraic identity $t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1t_2$, we substitute the expressions in terms of $k$:
$\dfrac{3h}{4} - 1 = \left(\dfrac{3k}{8} - 1\right)^2 - 2\left(-4 - \dfrac{3k}{8}\right)$
$\dfrac{3h}{4} - 1 = \dfrac{9k^2}{64} - \dfrac{3k}{4} + 1 + 8 + \dfrac{3k}{4}$
$\dfrac{3h}{4} - 1 = \dfrac{9k^2}{64} + 9$
$\dfrac{3h}{4} = \dfrac{9k^2}{64} + 10$
$3h = \dfrac{9k^2}{16} + 40$
$\dfrac{9k^2}{16} = 3h - 40$
$k^2 = \dfrac{16}{9}(3h - 40) = \dfrac{16}{3}\left(h - \dfrac{40}{3}\right)$

Replacing $(h, k)$ with $(x, y)$, the locus $C_o$ is the parabola:
$y^2 = \dfrac{16}{3}\left(x - \dfrac{40}{3}\right)$

The length of the latus rectum of this parabola is $\dfrac{16}{3}$.

Three times the length of the latus rectum is $3 \times \dfrac{16}{3} = 16$.

Answer: $16$