If a ne 6, b, c satisfy left| begin{array}{cc | vedclass

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(C) Given the determinant equation: $\left| \begin{array}{ccc} a & 2b & 2c \ 3 & b & c \ 4 & a & b \end{array} ight| = 0$. Applying the row operat

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$b^3$

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(C) Given the determinant equation: $\left| \begin{array}{ccc} a & 2b & 2c \ 3 & b & c \ 4 & a & b \end{array} ight| = 0$. Applying the row operation $R_1 	o R_1 - 2R_2$: $\left| \begin{array}{ccc} a - 6 & 2b - 2b & 2c - 2c \ 3 & b & c \ 4 & a & b \end{array} ight| = 0$ $\left| \begin{array}{ccc} a - 6 & 0 & 0 \ 3 & b & c \ 4 & a & b \end{array} ight| = 0$. Expanding along the first row: $(a - 6)(b^2 - ac) = 0$. Since $a 
e 6$,we must have $b^2 - ac = 0$,which implies $ac = b^2$. Multiplying both sides by $b$,we get $abc = b^3$.

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 = $

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