A) - 4
B) 2
C) 4
D) 8
Correct Answer: C
Solution :
\[\left| \,\begin{matrix} -{{a}^{2}} & ab & ac \\ ab & -{{b}^{2}} & bc \\ ac & bc & -{{c}^{2}} \\ \end{matrix} \right|=abc\left| \,\begin{matrix} -a & b & c \\ a & -b & c \\ a & b & -c \\ \end{matrix}\, \right|\] \[=(abc)(abc)\left| \,\begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \\ \end{matrix} \right|={{a}^{2}}{{b}^{2}}{{c}^{2}}(-1)(-4)\] \[=4{{a}^{2}}{{b}^{2}}{{c}^{2}}=K{{a}^{2}}{{b}^{2}}{{c}^{2}}\], (given) Þ K = 4.
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