8 vectors of given type are as follows

(1, 1, 1) $\underset{}{\overset{}{\leftrightarrow }}$  (–1, – 1, – 1)

(1, 1, – 1) $\underset{}{\overset{}{\leftrightarrow }}$  (–1, – 1, 1)

(1, – 1, 1) $\underset{}{\overset{}{\leftrightarrow }}$  (–1, 1, – 1)

(– 1, 1, 1) $\underset{}{\overset{}{\leftrightarrow }}$  (1, – 1, – 1)

The given pairs are collinear (anti parallel) any three pairs will be selected from the '4' available pairs and from each pair any one vector will be selected.

 $\therefore$        ⁴C₃ × ²C₁ × ²C₁ × ²C₁ = ⁴C₃ × 2³ = 2⁵ = 3²

i.e.        2^p = 32

             p = 5