Given equation of parabola is

$x^2-8x+12y+15=0$

$\Rightarrow x^2-8x+16=-12y-15+16$

$\Rightarrow {\left(x-4\right)}^2=-12\left(y-\frac{1}{12}\right)$

Compare it with ${\left(x-h\right)}^2=-4a\left(y-k\right)$

$\Rightarrow h=4,a=3, k=\frac{1}{12}$

As, we know that the parametric equation of ${\left(x-h\right)}^2=-4ay$ is $x=\mathrm{h}+2\mathrm{at},\mathrm{y}=\mathrm{k}-{\mathrm{at}}^2$.

$\therefore$Parametric equation of ${\left(x-4\right)}^2=-12\left(y-\frac{1}{12}\right)$ is

$x=4+6t ,y=\frac{1}{12}-3t^2$.

Hence, option $1$ is correct.