Step-by-step Solution

1. Understand the Set of Vectors V The set V is given by $V = \left\{ {a\,\hat i + b\,\hat j + c\hat k:a,\,b,\,c\, \in \left\{ { - 1,\,1} \right\}} \right\}$. The components a, b, and c can each take 2 values (-1 or 1). Therefore, the total number of vectors in the set V is 2 * 2 * 2 = 8. These 8 vectors represent the position vectors of the vertices of a cube centered at the origin with side length 2.

2. Formulate the Problem We need to find the number of ways to choose 3 non-coplanar vectors from the 8 vectors in V. Let this number be N. The problem states that $N = 2^p$, and we need to find the value of p.

3. Strategy: Total minus Unwanted A standard approach for this type of problem is to find the total number of ways to choose 3 vectors and then subtract the number of ways to choose 3 coplanar vectors.

   • Total number of ways to choose 3 vectors from 8 = C(8, 3).
   • Number of ways to choose 3 non-coplanar vectors = (Total ways) - (Number of ways to choose 3 coplanar vectors).

4. Calculate the Total Number of Selections The total number of ways to choose 3 vectors from the set of 8 vectors is: $C(8, 3) = \frac{{8!}}{{3!(8 - 3)!}} = \frac{{8!}}{{3!5!}} = \frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}} = 56$

5. Count the Number of Coplanar Selections Three vectors v1, v2, v3 are coplanar if they lie on the same plane passing through the origin. Geometrically, for the given set V, the vectors are coplanar if the corresponding vertices of the cube lie on a plane that passes through the origin.

   Let's identify the planes that pass through the origin and contain at least 3 vertices of the cube. These are the diagonal planes of the cube. There are 6 such planes:

   • x - y = 0
   • x + y = 0
   • x - z = 0
   • x + z = 0
   • y - z = 0
   • y + z = 0

   Let's analyze one of these planes, for example, x - y = 0 (or x=y). The vectors from V that lie on this plane are those where the x and y components are equal. These are:

   • (1, 1, 1)
   • (1, 1, -1)
   • (-1, -1, 1)
   • (-1, -1, -1) So, there are 4 vectors from V on this plane. Each of the 6 diagonal planes contains exactly 4 vectors from V.

   Any set of 3 vectors chosen from the 4 vectors lying on a single plane will be coplanar. The number of ways to choose 3 vectors from these 4 is: $C(4, 3) = \frac{{4!}}{{3!1!}} = 4$

   Since there are 6 such distinct planes, and any set of 3 non-collinear vectors lies on a unique plane, the total number of combinations of 3 coplanar vectors is: $\text{Number of coplanar sets} = (\text{Number of planes}) \times (\text{Combinations per plane})$ $\text{Number of coplanar sets} = 6 \times 4 = 24$

6. Calculate the Number of Non-Coplanar Selections The number of ways to choose 3 non-coplanar vectors is: $N = (\text{Total ways}) - (\text{Number of coplanar ways})$ $N = 56 - 24 = 32$

7. Find the value of p The problem states that the number of ways to choose 3 non-coplanar vectors is $2^p$. We found this number to be 32. So, $2^p = 32$. Since $32 = 2^5$, we have $2^p = 2^5$. Therefore, p = 5.

Final Answer

The number of ways to choose three non-coplanar vectors from V is 32. We are given this number is $2^p$. Thus, $2^p = 32$, which implies p = 5.