Question

If $3\left(a^2+b^2+c^2+1\right)=2(a+b+c+a b+b c+c a)$ then the value of $\begin{vmatrix}a & b & c \\ b c & c a & a b \\ 1 & 2 & 2\end{vmatrix}$ is equal to

• A. $a+b+c$
• B. $a+b-2 c$
• C. $a b c$
• D. $4 a^2 b^2 c^2$

Answer 1

Answer: (B)

$\text { Clearly, }( a -1)^2+( b -1)^2+( c -1)^2+( a - b )^2+( b - c )^2+( c - a )^2=0$
$\Rightarrow a = b = c =1$
$\therefore $ Determinant $=\begin{vmatrix}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 2 & 2\end{vmatrix}=0$