A) \[abc\]
B) \[4abc\]
C) \[4{{a}^{2}}{{b}^{2}}{{c}^{2}}\]
D) \[{{a}^{2}}{{b}^{2}}{{c}^{2}}\]
Correct Answer: C
Solution :
\[\Delta =\left| \,\begin{matrix} {{b}^{2}}+{{c}^{2}} & {{a}^{2}} & {{a}^{2}} \\ {{b}^{2}} & {{c}^{2}}+{{a}^{2}} & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{matrix}\, \right|\] =\[-2\,\left| \,\begin{matrix} 0 & {{c}^{2}} & {{b}^{2}} \\ {{b}^{2}} & {{c}^{2}}+{{a}^{2}} & {{b}^{2}} \\ {{c}^{2}} & {{c}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{matrix}\, \right|\],by \[{{R}_{1}}\to {{R}_{1}}-({{R}_{2}}+R{{ & }_{3}})\] = \[-2\,\left| \,\begin{matrix} 0 & {{c}^{2}} & {{b}^{2}} \\ {{b}^{2}} & {{a}^{2}} & 0 \\ {{c}^{2}} & 0 & {{a}^{2}} \\ \end{matrix}\, \right|\], by \[\begin{align} & {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\ & {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\ \end{align}\] = \[-2\{-{{c}^{2}}({{b}^{2}}{{a}^{2}})+{{b}^{2}}(-{{c}^{2}}{{a}^{2}})\}=4{{a}^{2}}{{b}^{2}}{{c}^{2}}\]. Trick: Put \[\alpha \] so that the option give different values.
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