If  $a\ne 6,b,c$ satisfy $\left|\begin{array}{ccc}a& 2b& 2c\\ 3& b& c\\ 4& a& b\end{array}\right|=0$, then  $abc=$

$a(b^2-ac)-3(2b^2-2ac)+4(2bc-2bc)=0$

$\Rightarrow a{b}^2-a^{2}c-6b^2+6ac=0$

$\Rightarrow b^2(a-6)-ac(a-6)=0$

$\Rightarrow (b^2-ac)(a-6)=0$

$\Rightarrow a=6\left(\text{or}\right)b^2=ac$

$\text{but\hspace{0.17em}}a\ne 6\left(\text{given}\right)$

$\therefore b^2=ac$

$\therefore abc=b^3$