If $Δ_1=\begin{vmatrix}1 & 1 & 1\\a & b & c\\a^2 & b^2 & c^2\end{vmatrix}, Δ_2=\begin{vmatrix}1 & bc & a\\1 & ca & b\\1 & ab& c\end{vmatrix}$, then

$Δ_1+Δ_2=0$

The correct answer is option (2) : $Δ_1+Δ_2=0$

$Δ_2= \begin{vmatrix}1 & bc & a\\1 & ca & b\\1 & ab & c\end{vmatrix}$

$⇒Δ_2=\frac{1}{abc}\begin{vmatrix}a & abc & a^2\\b & abc & b^2\\c & abc & c^2\end{vmatrix}$        $\begin{matrix}Applying \, R_1→R_1(a) \\R_2→R_2(b)\\R_3→R_3(c)\end{matrix}$

$⇒Δ_2= \begin{vmatrix}a & 1 & a^2\\b & 1 & b^2\\c & 1 & c^2\end{vmatrix}$            [Taking abc common from $C_2$]

$⇒Δ_2= -\begin{vmatrix}1 & a & a^2\\1 & b & b^2\\1 & c & c^2\end{vmatrix}$           [Interchanging $C_2 $ and $C_2 $ ]

$⇒Δ_2= -\begin{vmatrix}1 & 1 & 1\\a & b & c\\a^2 & b^2 & c^2\end{vmatrix}=-Δ_1 ⇒Δ_1+Δ_2=0.$