If a≠6,b,c satisfy a2b2c3bc4ab=0, then abc=

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If $a \neq 6, b, c$ satisfy $|\begin{matrix} a & 2 b & 2 c \\ 3 & b & c \\ 4 & a & b \end{matrix}| = 0$ , then $a b c =$

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$a + b + c$

$b^{3}$

$a b + b c$

answer is C.

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Detailed Solution

$a (b^{2} - a c) - 3 (2 b^{2} - 2 a c) + 4 (2 b c - 2 b c) = 0$

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

$\Rightarrow b^{2} (a - 6) - a c (a - 6) = 0$

$\Rightarrow (b^{2} - a c) (a - 6) = 0$

$\Rightarrow a = 6 (or) b^{2} = a c$

$but a \neq 6 (given)$

$∴ b^{2} = a c$

$∴ a b c = b^{3}$

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