(a, b, c, d)

Since , and are unit vectors inclined at an angle θ .

Therefore,

| | = | | = | | = 1 and cos θ = . = .

Now, = α + β + γ ( × ) …(1)

⇒ . = α ( . ) + β ( . ) + γ { .( × )}

⇒ cos θ = α | | 2 [  . = 0, . ( × ) = 0] ⇒ cos θ = α

Similarly, by taking dot product on both sides of (1) by we get β = cos θ .

∴ α = β .

Again, = α + β + γ ( × )

| | 2 = α 2 | | 2 + β 2 | | 2 + γ 2 | × | 2 + 2 α β ( . )

+ 2 α γ { . ( × )} + 2 β γ { . ( × )}

⇒ 1 = α 2 + β 2 + γ 2 | × | 2 ⇒ 1 = 2 α 2 + γ 2

⇒ 1 = 2 α 2 + γ 2 ⇒ α 2 =

But α = β = cos θ . Therefore, 1 = 2 α 2 + γ 2 ⇒ γ 2 = 1 – 2 cos 2 θ = – cos 2 θ

∴ α 2 = β 2 = = .

Hence (A, B, C, D) are the correct answer.