A) \[25\]
B) \[0\]
C) \[27\]
D) \[9\]
Correct Answer: B , C
Solution :
\[\vec{a},\vec{b},\vec{c}\] non-zero coplanar vectors \[[2\vec{a}-\vec{b}\,\,3\vec{b}-\vec{c}\,4\vec{c}-\vec{a}]\] \[=(2\vec{a}-\vec{b}).[(3\vec{b}-\vec{c})\times (4\vec{c}-\vec{a})]\] \[=(2\vec{a}-\vec{b}).[(12\vec{b}\times \vec{c}-4\vec{c}\times \vec{c}-3\vec{b}\times \vec{a}+\vec{c}\times \vec{a}]\] \[=(2\vec{a}-\vec{b})[12\vec{b}\times \vec{c}-3\vec{b}\times \vec{a}+\vec{c}\times \vec{a}]\] \[(\because \,\,\vec{c}\times \vec{c}=0)\] \[=24\vec{a}.(\vec{b}\times \vec{c})-6\vec{a}.(\vec{b}\times \vec{a})+2\vec{a}.(\vec{c}\times \vec{a})\] \[-12\vec{b}.(\vec{b}\times \vec{c})+3\vec{b}.(\vec{b}\times \vec{a})-\vec{b}.(\vec{c}\times \vec{a})\] \[=24[\vec{a}\vec{b}\vec{c}]-6[\vec{a}\vec{b}\vec{a}]+2[\vec{a}\vec{c}\vec{a}]\] \[-12[\vec{b}\vec{b}\vec{c}]+3[\vec{b}\vec{b}\vec{a}]-[\vec{b}\vec{c}\vec{a}]\] \[(\because \,[\vec{a}\,\vec{b}\,\vec{a}]=[\vec{a}\,\vec{c}\,\vec{a}]=[\vec{b}\,\vec{b}\,\vec{c}]=[\vec{b}\,\vec{b}\,\vec{a}]=0\] \[=24\,[\vec{a}\,\vec{b}\,\vec{c}]-[\vec{b}\,\,\vec{c}\,\,\vec{a}]\] \[(\because \,\,[\vec{a}\,\,\vec{b}\,\,\vec{c}]=[\vec{b}\,\,\vec{c}\,\,\vec{a}])\] \[=24\,\,[\vec{a}\,\,\vec{b}\,\,\vec{c}]-[\vec{a}\,\,\vec{b}\,\,\vec{c}]\] \[=23\,\,[\vec{a}\,\,\vec{b}\,\,\vec{c}]\] Given \[\vec{a},\] \[\vec{b},\] \[\vec{c}\] are coplanar that is \[[\vec{a}\,\vec{b}\,\vec{c}]=0\]. Hence, \[23\times 0=0\]Solution :
\[27\]
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