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# If $\small\begin{vmatrix}a^2&b^2&c^2\\(a+1)^2&(b+1)^2&(c+1)^2\\(a-1)^2&(b-1)^2&(c-1)^2\end{vmatrix}$ = $k\small\begin{vmatrix}a^2&b^2&c^2\\a&b&c\\1&1&1&\end{vmatrix}$ then the value of $k$ is

$(a)\;1\qquad(b)\;2\qquad(c)\;3\qquad(d)\;4$

Let $\Delta=\begin{vmatrix}a^2&b^2&c^2\\(a+1)^2&(b+1)^2&(c+1)^2\\(a-1)^2&(b-1)^2&(c-1)^2\end{vmatrix}$
Apply $R_2\Rightarrow R_2-R_3$
$\begin{vmatrix}a^2&b^2&c^2\\4a&4b&4c\\(a-1)^2&(b-1)^2&(c-1)^2\end{vmatrix}$
$\Rightarrow 4\begin{vmatrix}a^2&b^2&c^2\\a&b&c\\(a-1)^2&(b-1)^2&(c-1)^2\end{vmatrix}$
Apply $R_3\rightarrow R_3-(R_1-2R_2)$
$\Rightarrow 4\begin{vmatrix}a^2&b^2&c^2\\a&b&c\\1&1&1\end{vmatrix}$
$\therefore k=4$
hence (d) is the correct answer.