If an equilateral ∆ is inscribed in a parabola ${\mathrm{y}}^2=12\mathrm{x}$ with one of the vertex being at the vertex of the parabola then its height is

Given parabola is $y^2=12x$

Let $OAB$ is an equilateral triangle inscribed  in given parabola

from the figure

$\tan 30°=\frac{l}{h}$

it implies that $h=l\sqrt{3}$

and the point $\left(h,\frac{h}{\sqrt{3}}\right)$ lies on the parabola $y^2=12x$

Hence,

   $\frac{h^2}{3}=12h\Rightarrow h=36$

Therefore, the height of the equailateral triangle inscribed in the parabola is $\boxed{36}$