A) 2
B) 3
C) 0
D) 1
Correct Answer: D
Solution :
\[\left[ \overline{a}\times \overline{b}\,\,\overline{b}\times \overline{c}\,\,\overline{c}\times \overline{a} \right]\]Let\[\overline{u}=\overline{b}\times \overline{c}\] \[=\left( \vec{a}\times \vec{b} \right).\left\{ \left( \vec{b}\times \overline{c} \right)\times \left( \overline{c}\times \vec{a} \right) \right\}\] \[=\left( \vec{a}\times \vec{b} \right).\left\{ \vec{u}\times (\overline{c}\times \overline{a}) \right\}\] \[=\left( \vec{a}\times b \right).\left\{ \left( \overline{u}\cdot \vec{a} \right)\vec{c}-(\overline{u}\cdot \overline{c})\overline{a} \right\}\] \[=\left( \vec{a}\times \overline{b} \right).\left\{ \left[ \overline{b}\,\overline{c}\,\overline{a} \right]\vec{c}-\left[ \overline{b}\overline{c}\overline{c} \right]\vec{a} \right\}\] \[=\left[ \overline{b}\,\overline{c}\,\overline{a} \right]\left[ \vec{a}\,\overline{b}\,\overline{c} \right]\] \[={{\left[ \overline{b}\,\overline{c}\,\overline{a} \right]}^{2}}\]\[\Rightarrow \]\[\lambda =1\]
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