A) \[-2\]
B) 1
C) 2
D) \[abc\]
E) \[-1\]
Correct Answer: B
Solution :
\[\Delta =\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 1 & b & {{b}^{2}} \\ 1 & c & {{c}^{2}} \\ \end{matrix} \right|=\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 0 & b-a & {{b}^{2}}-{{a}^{2}} \\ 0 & c-a & {{c}^{2}}-{{a}^{2}} \\ \end{matrix} \right|\] \[=(b-a)(c-a)\left| \begin{matrix} 1 & a & {{a}^{2}} \\ 0 & 1 & b+a \\ 0 & c-a & c+a \\ \end{matrix} \right|\] \[=(b-a)(c-a)(c+a-b-a)\] \[\Rightarrow \]\[(b-a)(c-a)(c-b)\] \[\Rightarrow \]\[k(a-b)(b-c)(c-a)\] (given)
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