Given that,
- $f(x) = x^{2}-7x$
- $g(x) = x+3$
Now, $f(g(x))-3x = f(x+3)-3x = (x+3)^{2} – 7(x+3)-3x$
$\qquad \qquad \qquad = x^{2}+9+6x-7x-21-3x = x^{2}-4x-12$
Let $h(x) = x^{2}-4x-12$
$\Rightarrow h’(x) = 2x-4$
For minimum value of $x:$
$\Rightarrow h’(x) = 0.$
$\Rightarrow 2x-4=0$
$\Rightarrow \boxed{x=2}$
Now, $h(2) = 2^{2}-4(2)-12$
$\Rightarrow h(2)=4-8-12.$
$\Rightarrow \boxed{h(2) = -16}$
$\therefore$ The minimum possible value of $f(g(x))-3x$ is $-16.$
Correct Answer $:\text{A}$