Evaluate :

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

When $a=b$, the first two rows become identical. Hence, $a-b$ is a factor.

Similarly, when $b=c$ and $c=a$, the second and third and third and first rows become identical. Hence, $b-c$ and $c-a$ are also factors

The degree of product of the diagonal elements is 3 . Hence, there are no other factors.

$\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|=\lambda(a-b)(b-c)(c-a) \quad$ [Where $\lambda$ is a constant]

$\left|\begin{array}{lll}1 & 0 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 0\end{array}\right|=2 \lambda \quad[$ Putting $a=0, b=1$ and $c=2$ to find $\lambda]$

$\Rightarrow 2=2 \lambda$

$\Rightarrow \lambda=1$

$\Rightarrow 2=2 \lambda$

$\Rightarrow \lambda=1$

Hence,

$\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|=(a-b)(b-c)(c-a)$