A) \[a+b+c\]
B) \[abc\]
C) 4
D) 1
Correct Answer: C
Solution :
Let\[\Delta =\left| \begin{matrix} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & b+c & -2c \\ \end{matrix} \right|\] Applying \[{{C}_{1}}+{{C}_{3}}\]and\[{{C}_{2}}+{{C}_{3}}\] \[\Delta =\left| \begin{matrix} -a+c & 2a+b+c & a+c \\ 2b+a+c & -b+c & b+c \\ a-c & b-c & -2c \\ \end{matrix} \right|\] Now, applying \[{{R}_{1}}+{{R}_{3}}\]and\[{{R}_{2}}+{{R}_{3}}\] \[\Delta =\left| \begin{matrix} 0 & 2\left( a+b \right) & a-c \\ 2\left( a+b \right) & 0 & b-c \\ a-c & b-c & -2c \\ \end{matrix} \right|\] On expanding, we get \[\Delta =-2(a+b)\{-2c\}[2(a+b)]\] \[-(a-c)(b-c)\}\] \[+(a-c)[2(a+b)(b-c)]\] \[\Delta =8c(a+b)(a+b)\] \[+4(a+b)(a-c)(b-c)\] \[=4(a+b)[2ac+2bc+ab-bc-ac+{{c}^{2}}]\] \[=4(a+b)[ac+bc+ab+{{c}^{2}}]\] \[=4(a+b)[c(a+c)+b(a+c)]\] \[=4(a+b)(b+c)(c+a)\] \[=\alpha (a+b)(b+c)(c+a)\] Hence,\[\alpha =4\]
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