A) \[\left| \Delta \right|\]=3
B) \[\left| \Delta \right|\]=2
C) \[\left| \Delta \right|\]=1
D) \[\Delta \]=0
Correct Answer: C
Solution :
[c] we have \[{{\Delta }^{2}}=\Delta \Delta =\left| \begin{matrix} {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ {{l}_{3}} & {{m}_{3}} & {{n}_{3}} \\ \end{matrix} \right|\]\[\left| \begin{matrix} {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ {{l}_{3}} & {{m}_{3}} & {{n}_{3}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} {{l}_{1}}^{2}+m_{1}^{2}+n_{1}^{2} & {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}} & {{l}_{1}}{{l}_{3}}+{{m}_{1}}{{m}_{3}}+{{n}_{1}}{{n}_{3}} \\ {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}} & l_{2}^{2}+m_{2}^{2}+n_{2}^{2} & {{l}_{2}}{{l}_{3}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}} \\ {{l}_{1}}{{l}_{3}}+{{m}_{1}}{{m}_{3}}+{{n}_{1}}{{n}_{3}} & {{l}_{2}}{{l}_{3}}+{{m}_{2}}{{m}_{3}}+{{n}_{2}}{{n}_{3}} & l_{3}^{2}+m_{3}^{2}+n_{3}^{2} \\ \end{matrix} \right|\]\[=\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right|1\] \[\Rightarrow \Delta =\pm 1\Rightarrow \left| \Delta \right|=1\]
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