Question

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. $a+b+c$
• B. $0$
• C. $b^3$
• D. $ab+b-c$

Answer 1

The correct option is C $b^3$

Expanding the Determinant we get,

$a(b^2-ac)-2b(3b-4c)+2c(3a-4b)=0\Rightarrow a{b}^2-a^{2}c-6b^2+8bc+6ca-8bc=0\Rightarrow a{b}^2-6b^2=a^{2}c-6ca\Rightarrow (a-6)b^2=(a-6)ca\Rightarrow b^2=ca$

Multilpying $b$ both sides we get

$b^3=abc$