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If $a, b, c$ are three complex numbers such that $a^2 + b^2 + c^2 = 0$ and $\left| {\begin{array}{*{20}{c}}
{\left( {{b^2} + {c^2}} \right)}&{ab}&{ac}\\
{ab}&{\left( {{c^2} + {a^2}} \right)}&{bc}\\
{ac}&{bc}&{\left( {{a^2} + {b^2}} \right)}
\end{array}} \right| = K{a^2}{b^2}{c^2}$ then value of $K$ is
$1$
$2$
$-2$
$4$
Solution
${{\rm{C}}_1} \to {\rm{a}}{{\rm{C}}_1},{{\rm{C}}_2} \to {\rm{b}}{{\rm{C}}_2}$ and ${\rm{c}}{{\rm{C}}_3} \to {\rm{c}}$
$\frac{1}{a b c}\left|\begin{array}{ccc}{a\left(b^{2}+c^{2}\right)} & {a b^{2}} & {a c^{2}} \\ {a^{2} b} & {b\left(c^{2}+a^{2}\right)} & {b c^{2}} \\ {a^{2} c} & {b^{2} c} & {c\left(a^{2}+b^{2}\right)}\end{array}\right|$
$=\left|\begin{array}{ccc}{b^{2}+c^{2}} & {b^{2}} & {c^{2}} \\ {a^{2}} & {c^{2}+a^{2}} & {c^{2}} \\ {a^{2}} & {b^{2}} & {a^{2}+b^{2}}\end{array}\right|$
Use $\mathrm{C}_{1}>\left(\mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}\right)$ and solve
