Mathematics - Determinant Question with Solution | TestHub

We have,

$\left(1-a\right)x+y+z=0$

$x+\left(1-b\right)y+z=0$

$x+y+\left(1-c\right)z=0$

has many solutions 

$\therefore \left|\begin{array}{ccc}1-a& 1& 1\\ 1& 1-b& 1\\ 1& 1& 1-c\end{array}\right|=0$

$\Rightarrow ab+ac+bc=abc$

$$\begin{gathered} \therefore \frac{ab+bc+ac}{3}\ge {\left(a^2{b}^2{c}^2\right)}^{1/3} \\ \Rightarrow {\left(abc\right)}^{1/3}\ge 3 \end{gathered}$$

$\therefore$ Minimum of $abc={\left(3\right)}^3=27$