If a, b, c are non zero complex numbers satisfying a^2+b^2+c^2 =0 and…

The correct answer is option (1) : 4

Let $Δ=\begin{vmatrix}b^2+c^2 & ab & ac\\ab & c^2+a^2 & bc\\ac& bc & a^2+b^2\end{vmatrix}$

Applying $R_1→R_1(a), R_1 → R_2 (b) $ and $R_3 → R_3 (c),$ we get

$Δ=\frac{1}{abc}\begin{vmatrix}a(b^2+c^2) & a^2b & a^2c\\ab^2 & b(c^2+a^2) & b^2c\\ac^2& bc^2 & c(a^2+b^2)\end{vmatrix}$

Taking $a, b $ and $c$ common from $C_1, C_2 $ and $C_3 $ respectively, we get

$Δ=\frac{abc}{abc}\begin{vmatrix}b^2+c^2 & a^2 & a^2\\b^2 & c^2+a^2 & b^2\\c^2& c^2 & a^2+b^2\end{vmatrix}$

$⇒Δ=\begin{vmatrix}2(b^2+c^2) & 2(c^2+a^2) & 2(a^2+b^2)\\b^2 & c^2+a^2 & b^2\\c^2& c^2 & a^2+b^2\end{vmatrix}$

$⇒Δ=2\begin{vmatrix}b^2+c^2 & c^2+a^2 & a^2+b^2\\b^2 & c^2+a^2 & b^2\\c^2& c^2 & a^2+b^2\end{vmatrix}$   [Taking 2 common form $R_1 $ ]

Applying $R_2→R_2-R_1 $ and $R_3 → R_3 -R_1, $ we get

$Δ=2\begin{vmatrix}b^2+c^2 & c^2+a^2 & a^2+b^2\\-c^2 & 0 & -a^2\\-b^2& -a^2 & 0\end{vmatrix}$

$⇒Δ=2\begin{vmatrix}0 & c^2 & b^2\\-c^2 & 0 & -a^2\\-b^2& -a^2 & 0\end{vmatrix}$   Applying $R_1→R_1+R_2+R_3$

$⇒Δ=2\begin{Bmatrix}0\begin{vmatrix} 0 & -a^2\\-a^2 & 0\end{vmatrix}-c^2\begin{vmatrix} -c^2 & -a^2\\-b^2 & 0\end{vmatrix}+b^2\begin{vmatrix} -c^2 & 0\\-b^2 & -a^2\end{vmatrix}\end{Bmatrix}$

$⇒Δ=2 (a^2b^2c^2 +a^2b^2c^2 )= 4a^2b^2c^2 $

But, it is given that $Δ=ka^2b^2c^2$

$∴k= 4.$