Consider the set of eight vectors $V=\{a\widehat{i}+b\widehat{j}+c\widehat{k}:a,b,c\in \{-1,1\left\}\right\}.$Three non-coplanar vectors can be chosen from $V$ in $2^n$ ways. Then, n is

 

Eight vectors in$V$ are

$\begin{array}{l}\overrightarrow{u}=\widehat{i}+\widehat{j}+\widehat{k},\overrightarrow{v}=\widehat{i}+\widehat{j}-\widehat{k},\overrightarrow{w}=\widehat{i}-\widehat{j}+\widehat{k},\overrightarrow{\alpha }=\widehat{i}-\widehat{j}-\widehat{k}\\ {\to }_{u^{\mathrm{\prime }}}=\widehat{i}-\widehat{j}-\widehat{k},{\overrightarrow{v}}^{\mathrm{\prime }}=-\widehat{i}-\widehat{j}+\widehat{k},{\overrightarrow{w}}^{\mathrm{\prime }}=-\widehat{i}-\widehat{j}+\widehat{k},{\overrightarrow{\alpha }}^{\mathrm{\prime }}=-\widehat{i}+\widehat{j}+\widehat{k}\end{array}$

Clearly, $\overrightarrow{u},\overrightarrow{u^{\mathrm{\prime }}};\overrightarrow{v},\overrightarrow{v^{\mathrm{\prime }}};\overrightarrow{w},\overrightarrow{w^{\mathrm{\prime }}}$ and $\overrightarrow{\alpha },\overrightarrow{{\alpha }^{\mathrm{\prime }}}$are four pairs of unlike parallel vectors. If we take any one pair out of these four pairs, nd one vector from the remaining $6$ vectors, we obtain three coplanar vectors. Total number of ways of selecting three coplanar vectors

${=}^6{C}_1{\times }^4{C}_1=24$

Jso,

Number of ways of selecting $3$ vectors out of given $8$ vectors $= 8C3$ Number of ways of selecting 3 non-coplanar vectors 

${=}^8{C}_3-24=32=2^5$

Hence, n $=5.$