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 $abc=$

Given the determinant is zero:
$$\left|\begin{array}{ccc}a& 2b& 2c\\ 3& b& c\\ 4& a& b\end{array}\right|=0$$
Apply $R_1\to R_1-2R_2$:
$$\left|\begin{array}{ccc}a-6& 0& 0\\ 3& b& c\\ 4& a& b\end{array}\right|=0$$
Expanding along $R_1$:
$$(a-6)\cdot (b^2-ac)=0$$
Since $a\ne 6$, we must have $b^2-ac=0⟹ac=b^2$.Multiply both sides by $b$:
$$abc=b\cdot b^2=b^3$$