Selection of Non Coplanar Vectors from Cube Vertices

To find the number of ways to choose three non-coplanar vectors from the set $V = \{a\hat{i} + b\hat{j} + c\hat{k} : a, b, c \in \{-1, 1\}\}$, we can first determine the total number of ways to choose any three vectors and then subtract the number of ways to choose three coplanar vectors.

Step 1: Total number of ways to choose 3 vectors

The set $V$ contains $2^3 = 8$ distinct vectors. The total number of ways to choose 3 vectors from these 8 vectors is: $\text{Total ways} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Step 2: Finding the condition for coplanarity

Three vectors $\vec{u}, \vec{v}, \vec{w} \in V$ are coplanar if and only if their scalar triple product is zero, which is equivalent to:

$$\det \begin{pmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{pmatrix} = 0$$

where all components are $\pm 1$.

Since the entries of each vector belong to $\{-1, 1\}$, any linear combination of two non-collinear vectors $\vec{u}$ and $\vec{v}$ that results in a vector $\vec{w}$ with components in $\{-1, 1\}$ requires one of the coefficients to be zero.

Thus, three vectors from $V$ are linearly dependent (coplanar) if and only if at least two of the chosen vectors are collinear. Since all vectors in $V$ have the same magnitude $\sqrt{3}$, two distinct collinear vectors must be exact opposites of each other (i.e., $\vec{u} = -\vec{v}$).

Step 3: Counting the number of coplanar sets

The set $V$ can be partitioned into 4 pairs of diametrically opposite vectors:

1. $(1, 1, 1)$ and $(-1, -1, -1)$
2. $(1, 1, -1)$ and $(-1, -1, 1)$
3. $(1, -1, 1)$ and $(-1, 1, -1)$
4. $(-1, 1, 1)$ and $(1, -1, -1)$

To form a set of three coplanar vectors:

1. Choose $1$ opposite pair from the $4$ available pairs: $\binom{4}{1} = 4$ ways.
2. Choose any $1$ third vector from the remaining $8 - 2 = 6$ vectors: $\binom{6}{1} = 6$ ways.

Therefore, the number of coplanar triplets is: $\text{Number of coplanar triplets} = 4 \times 6 = 24$

Step 4: Finding the number of non-coplanar triplets

$\text{Number of non-coplanar triplets} = \text{Total triplets} - \text{Coplanar triplets} = 56 - 24 = 32$

We are given that the number of non-coplanar ways is $2^p$: $2^p = 32 = 2^5$

Thus, $p = 5$