If a, b, c are three non-zero, non-coplanar vectors and \[{b_1} = b - \dfrac{{b \cdot a}}{{|a{|^2}}}a,{b_2} = b + \dfrac{{b \cdot a}}{{|a{|^2}}}a,{c_1} = c - \dfrac{{c.a}}{{|a{|^2}}}a - \dfrac{{c.b}}{{|b{|^2}}}b,{c_2} = c - \dfrac{{c.a}}{{|a{|^2}}}a\dfrac{{c.{b_1}}}{{{{\left| {{b_1}} \right|}^2}}}{b_1}\]\[{c_3} = c - \dfrac{{c.a}}{{|a{|^2}}}a\dfrac{{c.{b_2}}}{{{{\left| {{b_2}} \right|}^2}}}{b_2}\mid {c_4} = a - \dfrac{{c.a}}{{|a{|^2}}}a\]
Then which of the following is a set of mutually orthogonal vectors is
A. \[\left\{ {{\bf{a}},{{\bf{b}}_1},{c_1}} \right\}\]
B. \[\left\{ {{\bf{a}},{{\bf{b}}_1},{c_2}} \right\}\]
C. \[\left\{ {{\bf{a}},{{\bf{b}}_2},{c_3}} \right\}\]
D. \[\left\{ {{\bf{a}},{{\bf{b}}_2},{c_4}} \right\}\]

Hint: The idea of orthogonal vectors and its definition are used to determine whether two vectors are orthogonal to one other by computing the dot product between the vectors. Similarly, the idea of parallel vectors is used, where the magnitude and angle between them are determined in order to determine whether the given vectors are parallel. In this case, we have to replace the given value in the corresponding equation in order to determine the desired answer.

Formula used: Vectors \[\vec a,\vec b,\vec c\] are coplanar if their scalar triple product is zero.
\[[\vec a\vec b\vec c] = 0\]

Complete step by step solution: We have given in the problem that, a, b, c are three non-zero, non-coplanar vectors.
And it is also given in the question that,
\[{b_1} = b - \dfrac{{b \cdot a}}{{|a{|^2}}}a,{b_2} = b + \dfrac{{b \cdot a}}{{|a{|^2}}}a,{c_1} = c - \dfrac{{c.a}}{{|a{|^2}}}a - \dfrac{{c.b}}{{|b{|^2}}}b,{c_2} = c - \dfrac{{c.a}}{{|a{|^2}}}a\dfrac{{c.{b_1}}}{{{{\left| {{b_1}} \right|}^2}}}{b_1}\]
\[{c_3} = c - \dfrac{{c.a}}{{|a{|^2}}}a\dfrac{{c.{b_2}}}{{{{\left| {{b_2}} \right|}^2}}}{b_2}\mid {c_4} = a - \dfrac{{c.a}}{{|a{|^2}}}a\]
We have already known that the pair of vectors \[(\vec a,\vec b)\] is orthogonal if their product is zero.
That is, the mathematical format of the above expression can be written as,
\[\vec a \cdot \vec b = 0\]
Now, let us consider according to the given question that,
\[\vec a \cdot {\vec b_1} = \left( {\vec b \cdot \vec a - \dfrac{{\vec b \cdot \vec a}}{{|\vec a{|^2}}}\vec a \cdot \vec a} \right)\]
Now, we have to simplify the above expression.
Since, it is \[|\vec a{|^2} = 1,\vec a \cdot \vec a = 1\]we get
\[ = \vec b \cdot \vec a - \vec b \cdot \vec a\]
Now, let’s simplify the above expression, we obtain (as both are similar terms, it get cancelled)
\[ = 0\]
Now, we have to consider the below as per the given question, we get\[{\vec c_2} \cdot \vec a = \vec c \cdot \vec a - \dfrac{{\vec c \cdot \vec a}}{{|\vec a{|^2}}}\vec a \cdot \vec a - \dfrac{{{{\vec b}_1} \cdot \vec c}}{{{{\left| {\overrightarrow {{b_1}} } \right|}^2}}} \cdot \vec a\]
Now, let’s simplify the obtained expression, we get
\[ = \vec c \cdot \vec a - \vec c \cdot \vec a - 0 \ldots .\]
Since, it is\[\left( {|\vec a{|^2} = 1,\vec a \cdot \vec a = 1,\vec a \cdot {{\vec b}_1} = 0} \right)\]
Similarly, after solving it becomes
\[{\overrightarrow {\rm{b}} _1} \cdot {\overrightarrow {\rm{c}} _2} = 0\]
Now, \[\vec a,\overrightarrow {{b_1}} ,\overrightarrow {{c_2}} \] are the set of orthogonal vectors
Therefore, if a, b, c are three non-zero, non-coplanar vectors then the set of mutually orthogonal vectors is \[\vec a,\overrightarrow {{b_1}} ,\overrightarrow {{c_2}} \]

Thus, Option (B) is correct.

Note: The cross product of two vectors denotes a vector that is perpendicular to both of them. And you may find its unit vector by dividing the resultant vector by its magnitude. A common error is that the formula for determining orthogonality or the method used for determining the dot product are not used.