A) 0
B) 1
C) \[l+m+n+2\]
D) \[2(Z+m+n)+3\]
E) \[lmn-1\]
Correct Answer: B
Solution :
\[\left| \begin{matrix} 1+{{l}^{2}} & lm & \ln \\ lm & 1+{{m}^{2}} & mn \\ \ln & mn & 1+{{n}^{2}} \\ \end{matrix} \right|\] \[=(1+{{l}^{2}})\left| \begin{matrix} 1+{{m}^{2}} & mn \\ mn & 1+{{n}^{2}} \\ \end{matrix} \right|-lm\left| \begin{matrix} lm & mn \\ \ln & 1+{{n}^{2}} \\ \end{matrix} \right|\] \[=ln\left| \begin{matrix} lm & 1+{{m}^{2}} \\ ln & mn \\ \end{matrix} \right|\] \[=(1+{{l}^{2}})(1+{{m}^{2}}+{{n}^{2}}+{{m}^{2}}{{n}^{2}}-{{m}^{2}}{{n}^{2}})\] \[-lm(lm-lm{{n}^{2}}-lm{{n}^{2}})\] \[+\ln (l{{m}^{2}}n-\ln -l{{m}^{2}}n)\] \[=(1+{{l}^{2}})(1+{{m}^{2}}+{{n}^{2}})-{{l}^{2}}{{m}^{2}}-{{l}^{2}}{{n}^{2}}\] \[=1+{{m}^{2}}+{{n}^{2}}+{{l}^{2}}+{{l}^{2}}{{m}^{2}}+{{l}^{2}}{{n}^{2}}-{{l}^{2}}{{m}^{2}}-{{l}^{2}}{{n}^{2}}\] \[=1+{{m}^{2}}+{{n}^{2}}+{{l}^{2}}=1\] \[(\because {{l}^{2}}+{{m}^{2}}+{{n}^{2}}=0)\]
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