MCQ
If $a \ne 6,b,c$ satisfy $\left| {\,\begin{array}{*{20}{c}}a&{2b}&{2c}\\3&b&c\\4&a&b\end{array}\,} \right| = 0,$then $abc = $
- A$a + b + c$
- B$0$
- ✓${b^3}$
- D$ab + bc$
==> $(a - 6)({b^2} - ac) = 0 \Rightarrow {b^2} - ac = 0$ $a \ne 6$
$\therefore $ $ac = {b^2} \Rightarrow abc = {b^3}.$
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