A) 0
B) \[\alpha \text{ }\mathbf{a}\]
C) \[\beta \text{ }\mathbf{b}\]
D) \[(\alpha +\beta )\,\mathbf{c}\]
Correct Answer: A
Solution :
We have \[\mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \text{ }\mathbf{d}\] and \[\mathbf{b}+\mathbf{c}+\mathbf{d}=\beta \,\mathbf{a}\] \[\therefore \,\,\,\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=(\alpha +1)\mathbf{d}\]and \[\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=(\beta +1)\,\mathbf{a}.\] \[\Rightarrow (\alpha +1)\,\mathbf{d}=(\beta +1)\,\mathbf{a}\] If \[\alpha \,\ne \,-1,\] then \[(\alpha +1)\,\mathbf{d}=(\beta +1)\,\mathbf{a}\Rightarrow \mathbf{d}=\frac{\beta +1}{\alpha +1}\mathbf{a}\] \[\Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \,\mathbf{d}\Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c}=\alpha \left( \frac{\beta +1}{\alpha +1} \right)\mathbf{a}\] \[\Rightarrow \left( 1-\frac{\alpha (\beta +1)}{\alpha +1} \right)\,\mathbf{a}+\mathbf{b}+\mathbf{c}=0\] \[\Rightarrow \mathbf{a},\,\mathbf{b},\,\mathbf{c}\] are coplanar which is contradiction to the given condition, \[\therefore \,\,\,\alpha =-1\] and so \[\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=0.\]
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