If $\vec{a},\vec{b},\vec{c}$are three non-zero, non-coplanar vectrors and $\overrightarrow {{b_1}} \, = \,\overrightarrow {b\,} \, - \,\frac{{\overrightarrow b \,.\,\overrightarrow a }}{{{{\left| {\overrightarrow a \,} \right|}^2}}}\overrightarrow a \,,\,\overrightarrow {{b_2}} \, = \overrightarrow b \, + \,\frac{{\overrightarrow b \,.\,\overrightarrow a }}{{{{\left| {\overrightarrow a \,} \right|}^2}}}\overrightarrow a \, $ and $ \overrightarrow {{c_1}} \, = \,\overrightarrow c \, - \,\frac{{\overrightarrow c \,.\,\overrightarrow a }}{{{{\left| {\overrightarrow a \,} \right|}^2}}}\overrightarrow a \, + \,\frac{{\overrightarrow c \,.\,\overrightarrow b }}{{{{\left| {\overrightarrow b \,} \right|}^2}}}\overrightarrow {{b_1}} \, $, $\overrightarrow {{c_2}} \, = \,\overrightarrow c \, - \,\frac{{\overrightarrow c \,.\,\overrightarrow a }}{{{{\left| {\overrightarrow a \,} \right|}^2}}}\overrightarrow a \, - \,\frac{{\overrightarrow c \,.\,\overrightarrow b }}{{{{\left| {\overrightarrow {{b_1}} \,} \right|}^2}}}\overrightarrow {{b_1}} \, ,$ $ \overrightarrow {{c_3}} \, = \,\overrightarrow c \, - \,\frac{{\overrightarrow c \,.\,\overrightarrow a }}{{{{\left| {\overrightarrow c \,} \right|}^2}}}\overrightarrow a \, + \,\frac{{\overrightarrow c \,.\,\overrightarrow {{b_2}} }}{{{{\left| {\overrightarrow c \,} \right|}^2}}}\overrightarrow {{b_1}} \, $ $, \overrightarrow {{c_4}} \, = \,\overrightarrow c \, - \,\frac{{\overrightarrow c \,.\,\overrightarrow a }}{{{{\left| {\overrightarrow c \,} \right|}^2}}}\overrightarrow a \, - \,\frac{{\overrightarrow b \,.\,\overrightarrow c }}{{{{\left| {\overrightarrow b \,} \right|}^2}}}\overrightarrow {{b_1}} \,.$ Then, which of the following is a set of mutually orthogonal vectors ?
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