A) \[\log \,(abc)\]
B) \[1-\log \,\,abc\]
C) \[\log \,(a+b+c)\]
D) \[1+\,\log \,abc\]
Correct Answer: D
Solution :
\[\left| \begin{matrix} 1+\log a & \log \,b & \log \,c \\ \log \,a & 1+\log \,b & \log c \\ \log \,a & \log \,b & 1+\log c \\ \end{matrix} \right|\] Using the operation \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}},\] we get \[\left| \begin{matrix} 1+\log a+\log b+\log c & \log b & \log c \\ 1+\log a+\log b+\log c & 1+\log b & \log c \\ 1+\log a+\log b+\log c & \log b & 1+\log c \\ \end{matrix} \right|\] \[=(1+\log a+\log b+\log c)\] \[\left| \begin{matrix} 1 & \log b & \log c \\ 1 & 1+\log b & \log c \\ 1 & \log b & 1+\log c \\ \end{matrix} \right|\] \[=(1+\log a+\log b+\log c)\left| \begin{matrix} 1 & \log b & \operatorname{logc} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right|\] \[({{R}_{2}}\to {{R}_{2}}-{{R}_{1}},\,{{R}_{3}}\to {{R}_{3}}-{{R}_{1}})\] \[=(1+\log a+\log b+\log c).1\] \[=1+\log \,abc\]
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