Question

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

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

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 $=\left|\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 2& 2\end{array}\right|=0$