If \[\mathbf{d}=\lambda \,(\mathbf{a}\times \mathbf{b})+\mu \,(\mathbf{b}\times \mathbf{c})+\nu \,(\mathbf{c}\times \mathbf{a})\]and \[[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=\frac{1}{8},\] then \[\lambda +\mu +\nu \] is equal to

\[8\mathbf{d}\,\times \,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]

\[8\mathbf{d}\,.\,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]

\[\frac{\mathbf{d}\,}{8}.\,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]

\[\frac{\mathbf{d}\,}{8}\times \,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]

If \[\mathbf{d}=\lambda \,(\mathbf{a}\times \mathbf{b})+\mu \,(\mathbf

1.	If \[\mathbf{d}=\lambda \,(\mathbf{a}\times \mathbf{b})+\mu \,(\mathbf{b}\times \mathbf{c})+\nu \,(\mathbf{c}\times \mathbf{a})\]and \[[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=\frac{1}{8},\] then \[\lambda +\mu +\nu \] is equal to
A.	\[8\mathbf{d}\,.\,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]
B.	\[8\mathbf{d}\,\times \,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]
C.	\[\frac{\mathbf{d}\,}{8}.\,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]
D.	\[\frac{\mathbf{d}\,}{8}\times \,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]
Answer» B. \[8\mathbf{d}\,\times \,(\mathbf{a}+\mathbf{b}+\mathbf{c})\]

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If \[\mathbf{d}=\lambda \,(\mathbf{a}\times \mathbf{b})+\mu \,(\mathbf{b}\times \mathbf{c})+\nu \,(\mathbf{c}\times \mathbf{a})\]and \[[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]=\frac{1}{8},\] then \[\lambda +\mu +\nu \] is equal to

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