JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Vector Algebra

The set $V$ contains 8 vectors corresponding to the vertices of a cube centered at the origin. The total number of ways to choose 3 vectors from 8 is $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$. Three vectors are coplanar if and only if they include a pair of opposite vectors (negatives of each other). There are 4 such pairs: $(\hat{i}+\hat{j}+\hat{k},-\hat{i}-\hat{j}-\hat{k})$, $(\hat{i}+\hat{j}-\hat{k},-\hat{i}-\hat{j}+\hat{k})$, $(\hat{i}-\hat{j}+\hat{k},-\hat{i}+\hat{j}-\hat{k})$, $(-\hat{i}+\hat{j}+\hat{k},\hat{i}-\hat{j}-\hat{k})$. For each pair, choosing one vector from the pair and one from the remaining 6 gives $4\times 6=24$ coplanar triples. Thus, the number of non-coplanar triples is $56-24=32=2^5$, so $p=5$.