Let ∧u = u₁∧i + u₂∧j + u₃∧k be a unit vector in R³ and ∧w = 1/√6(∧i + ∧j + 2∧k). Given that there exists a vector →v in R³ such that ||∧u × →v|| = 1 and ∧w.(∧u × →v) = 1. Which of the following statements is(are) correct?

∧w.(∧u × →v) = 1 ⇒ |w||∧u × →v|cosα = 1 ⇒ cosα = 1
So, ∧w is parallel to (∧u × →v) ⇒ ∧w ⊥ ∧u and ∧w ⊥ →v
It is given that there exists a vector →v, So, there is a vector →v for every possible ∧u
Since ∧w ⊥ ∧u ⇒ ∧w.∧u = 0 ⇒ u₁ + u₂ + 2u₃ = 0
Infinitely many vectors satisfy this condition, so there are infinite choices for ∧u ⇒ there are infinitely many choices for →v
Now, If ∧u lies in the xy-plane ⇒ u₃ = 0 ⇒ |u₁| = |u₂|
If ∧u lies in the xz-plane ⇒ u₂ = 0 ⇒ |u₁| = 2|u₃|
So, Options (B, C) are correct