jmc

Combining denominators and simplifying,$$\frac{F(3x)}{F(x+3)}=\frac{9(x^2+5x+6)-48x-54}{x^2+5x+6}=\frac{9x^2-3x}{x^2+5x+6}=\frac{3x(3x-1)}{(x+3)(x+2)}$$It becomes obvious that $F(x)=ax(x-1)$, for some constant $a$, matches the definition of the polynomial. To prove that $F(x)$ must have this form, note that$$(x+3)(x+2)F(3x)=3x(3x-1)F(x+3)$$ Since $3x$ and $3x-1$ divides the right side of the equation, $3x$ and $3x-1$ divides the left side of the equation. Thus $3x(3x-1)$ divides $F(3x)$, so $x(x-1)$ divides $F(x)$. It is easy to see that $F(x)$ is a quadratic, thus $F(x)=ax(x-1)$ as desired. By the given, $F(6)=a(6)(5)=15⟹a=\frac{1}{2}$. Thus, $F(12)=\frac{1}{2}(12)(11)=\boxed{66}$.