Let →a=α1^i+α2^j+α3^k,→b=β1^i+β2^j+β3^k and →c=γ1^i+γ2^j+γ3^k, |→a|=2√2,→a makes an angle π6 with the plane of →b,→c and (→b,→c)=π3, then ∣∣∣∣α1α2α3β1β2β3γ1γ2γ3∣∣∣∣nis equal to(n is even natural number)

Let →a=α1^i+α2^j+α3^k,→b=β1^i+β2^j+β3^k and →c=γ1^i+γ2^j

1.	Let →a=α1^i+α2^j+α3^k,→b=β1^i+β2^j+β3^k and →c=γ1^i+γ2^j+γ3^k, \|→a\|=2√2,→a makes an angle π6 with the plane of →b,→c and (→b,→c)=π3, then ∣∣∣∣α1α2α3β1β2β3γ1γ2γ3∣∣∣∣nis equal to(n is even natural number)
Answer» $Let \to a = α_{1}^i + α_{2}^j + α_{3}^k,$ $\to b = β_{1}^i + β_{2}^j + β_{3}^k$ and $\to c = γ_{1}^i + γ_{2}^j + γ_{3}^k,$ $\| \to a \| = 2 \sqrt{2}, \to a$ makes an angle $\frac{π}{6}$ with the plane of $\to b, \to c$ and $(\to b, \to c) = \frac{π}{3},$ then ${∣ ∣∣ ∣ \begin{matrix} α_{1} & α_{2} & α_{3} β_{1} & β_{2} & β_{3} γ_{1} & γ_{2} & γ_{3} \end{matrix} ∣ ∣∣ ∣}^{n}$ is equal to ( $n$ is even natural number)