A) 1
B) 3
C) 4
D) 2
Correct Answer: C
Solution :
Let\[\Delta =\left| \begin{matrix} {{b}^{2}}+{{c}^{2}} & ab & ac \\ ab & {{c}^{2}}+{{a}^{2}} & bc \\ ac & bc & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right|\] Multiply \[{{C}_{1}}\] by a, \[{{C}_{2}}\] by b and \[{{C}_{3}}\] by c and hence divide by abc. \[=\frac{1}{abc}\left| \begin{matrix} a\left( {{b}^{2}}+{{c}^{2}} \right) & a{{b}^{2}} & a{{c}^{2}} \\ {{a}^{2}}b & b\left( {{c}^{2}}+{{a}^{2}} \right) & b{{c}^{2}} \\ {{a}^{2}}c & {{b}^{2}}c & c\left( {{a}^{2}}+{{b}^{2}} \right) \\ \end{matrix} \right|\] Take out a, b, c common from \[{{R}_{1}},{{R}_{2}}\] and \[{{R}_{3}}\]respectively. \[\therefore \]\[\Delta =\frac{abc}{abc}\left| \begin{matrix} {{b}^{2}}+{{c}^{2}} & {{b}^{2}} & {{c}^{2}} \\ {{a}^{2}} & {{c}^{2}}+{{a}^{2}} & {{c}^{2}} \\ {{a}^{2}} & {{b}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right|\] Apply \[{{C}_{1}}\to {{C}_{1}}-{{C}_{2}}-{{C}_{3}}\] \[\Delta =\left| \begin{matrix} 0 & {{b}^{2}} & {{c}^{2}} \\ -2{{a}^{2}} & {{c}^{2}}+{{a}^{2}} & {{c}^{2}} \\ -2{{b}^{2}} & {{b}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right|\] \[=-2\left| \begin{matrix} 0 & {{b}^{2}} & {{c}^{2}} \\ {{c}^{2}} & {{c}^{2}}+{{a}^{2}} & {{c}^{2}} \\ {{b}^{2}} & {{b}^{2}} & {{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right|\] Apply\[{{C}_{2}}-{{C}_{1}}\]and\[{{C}_{3}}-{{C}_{1}}\] \[=-2\left| \begin{matrix} 0 & {{b}^{2}} & {{c}^{2}} \\ {{c}^{2}} & {{a}^{2}} & 0 \\ {{b}^{2}} & 0 & {{a}^{2}} \\ \end{matrix} \right|\] \[=-2[-{{b}^{2}}({{c}^{2}}{{a}^{2}})+{{c}^{2}}(-{{a}^{2}}{{b}^{2}})]\] \[=2{{a}^{2}}{{b}^{2}}{{c}^{2}}+2{{a}^{2}}{{b}^{2}}{{c}^{2}}=4{{a}^{2}}{{b}^{2}}{{c}^{2}}\] But\[\Delta =k{{a}^{2}}{{b}^{2}}{{c}^{2}}\therefore k=4\]
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