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(D) Given $a^2 + b^2 + c^2 = 0$. We know that $b^2 + c^2 = -a^2$,$c^2 + a^2 = -b^2$,and $a^2 + b^2 = -c^2$.Substituting these into the determinant $\D

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$4$

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(D) Given $a^2 + b^2 + c^2 = 0$. We know that $b^2 + c^2 = -a^2$,$c^2 + a^2 = -b^2$,and $a^2 + b^2 = -c^2$.Substituting these into the determinant $\Delta$:$\Delta = \begin{vmatrix} -a^2 & ab & ac \ ab & -b^2 & bc \ ac & bc & -c^2 \end{vmatrix}$Taking $a$ common from $R_1$,$b$ from $R_2$,and $c$ from $R_3$:$\Delta = abc \begin{vmatrix} -a & b & c \ a & -b & c \ a & b & -c \end{vmatrix}$Taking $a$ common from $C_1$,$b$ from $C_2$,and $c$ from $C_3$:$\Delta = (abc)(abc) \begin{vmatrix} -1 & 1 & 1 \ 1 & -1 & 1 \ 1 & 1 & -1 \end{vmatrix}$$\Delta = a^2 b^2 c^2 \begin{vmatrix} -1 & 1 & 1 \ 1 & -1 & 1 \ 1 & 1 & -1 \end{vmatrix}$Evaluating the determinant:$-1(1 - 1) - 1(-1 - 1) + 1(1 - (-1)) = 0 + 2 + 2 = 4$Thus,$\Delta = 4 a^2 b^2 c^2$.Comparing with $K a^2 b^2 c^2$,we get $K = 4$.

If $a, b, c$ are three complex numbers such that $a^2 + b^2 + c^2 = 0$ and $\begin{vmatrix} (b^2 + c^2) & ab & ac \\ ab & (c^2 + a^2) & bc \\ ac & bc & (a^2 + b^2) \end{vmatrix} = K a^2 b^2 c^2$,then the value of $K$ is:

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