Animated Solution for Mathematics - Vector Algebra: Consider the set of eight vectors $V=\{a\hat{i}+b\hat{j}+c\hat{k}:a,b,c\in \{-1,1\}\}$. Three non-coplanar vectors can be chosen from $V$ in $2^p$ ways. Then $p$ is .........

The set $V=\{a\hat{i}+b\hat{j}+c\hat{k}:a,b,c\in \{-1,1\}\}$ represents the eight vertices of a cube centered at the origin. Since each component has two possible values, we confirm the total number of vectors is $2\times 2\times 2=8$.

These vectors exhibit central symmetry. For every vector $\vec{v}\in V$, its diametrically opposite vector $-\vec{v}$ is also in $V$. This partitions the set of eight vectors into four pairs of opposite vectors.

We are looking for the number of ways to choose three vectors from $V$ such that they are not coplanar. A set of three vectors is coplanar if they lie in the same plane passing through the origin.

If we select both vectors from any opposite pair, they are collinear. Any third vector chosen will then form a plane with this line, resulting in a coplanar triplet. Therefore, to ensure the three vectors are non-coplanar, we must select three vectors such that no two are opposite to each other.

To satisfy the condition, we must choose three distinct pairs out of the four available pairs. The number of ways to select these pairs is given by the combination formula:

Once we have selected three pairs, we must choose exactly one vector from each of the three chosen pairs. Since each pair contains two vectors, there are $2$ choices for each pair.

For a specific selection of three pairs, the number of ways to pick the vectors is:

To find the total number of non-coplanar triplets, we multiply the number of ways to choose the pairs by the number of ways to choose the vectors within those pairs:

Given that the problem states the answer is in the form $2^p$, we set:

Thus, we find that $p=5$.