If $\Delta_{1}=\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2}\end{array}\right|, \Delta_{2}=\left|\begin{array}{ccc}1 & b c & a \\ 1 & c a & b \\ 1 & a b & c\end{array}\right|$, then
(a) $\Delta_{1}+\Delta_{2}=0$
(b) $\Delta_{1}+2 \Delta_{2}=0$
(c) $\Delta_{1}=\Delta_{2}$
(d) none of these
(a) $\Delta_{1}+\Delta_{2}=0$
$\Delta_{2}=\mid \begin{array}{lll}1 & b c & a\end{array}$
$1 \quad c a \quad b$
$\begin{array}{lll}1 & a b & c\end{array}$
$=\frac{1}{a b c} \mid \begin{array}{lll}a & a b c & a^{2}\end{array}$
$\begin{array}{lcc}c & c a b & c^{2} \mid\end{array}\left[R_{1}, R_{2}, R_{3}\right.$ are multiplied by a, b and c respectively, therefore we divide by abc]
$=\frac{a b c}{a b c} \mid a \quad 1 \quad a^{2}$
$b \quad 1 \quad b^{2}$
$c \quad 1 \quad c^{2}$ [Taking $a b c$ common from $C_{2}$ ]
$=-\mid \begin{array}{lll}1 & a & a^{2}\end{array}$
$1 \quad b \quad b^{2}$
$\begin{array}{llll}1 & c & c^{2} & C_{1} \leftrightarrow C_{2}\end{array}$
We know that the value of a determinant remains unchanged if its rows and columns are interchanged. So,
$\Delta_{2}=-\mid \begin{array}{lll}1 & 1 & 1\end{array}$
$a \quad b \quad c$
$\begin{array}{lll}a^{2} & b^{2} & c^{2}\end{array}$
$=-\Delta_{1}$
$\Rightarrow \Delta_{1}+\Delta_{2}=0$