A) \[abc\]
B) \[-(a+b+c)\]
C) \[-2(a+b+c)\]
D) \[2(a+b+c)\]
Correct Answer: C
Solution :
Here,\[\left| \begin{matrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \\ \end{matrix} \right|\] Applying \[{{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}},\]we get \[=\left| \begin{matrix} a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \\ \end{matrix} \right|\] \[=(a+b+c)\left| \begin{matrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \\ \end{matrix} \right|\] Applying\[{{C}_{2}}\to {{C}_{2}}-{{C}_{1}},{{C}_{3}}\to {{C}_{3}}-{{C}_{1}},\]we get \[=(a+b+c)\left| \begin{matrix} 1 & 0 & 0 \\ 2b & -(a+b+c) & 0 \\ 2c & 0 & -(a+b+c) \\ \end{matrix} \right|\] \[=(a+b+c){{(a+b+c)}^{2}}\] Now,\[=(a+b+c){{(a+b+c)}^{2}}=(a+b+c)\]\[{{(x+a+b+c)}^{2}}\] \[\Rightarrow \]\[x=-2(a+b+c)\]
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