Let →a=α^i+2^j−3^k,→b=^i+2α^j−2^k and →c=2^i−α^j+^k and {(→a×→b)×(→b×→c)}×(→c×→a)=→0, then the value of 6α is

Let →a=α^i+2^j−3^k,→b=^i+2α^j−2^k and →c=2^i−α^j+^k and

1.	Let →a=α^i+2^j−3^k,→b=^i+2α^j−2^k and →c=2^i−α^j+^k and {(→a×→b)×(→b×→c)}×(→c×→a)=→0, then the value of 6α is
Answer» Let $\to a = α^i + 2^j - 3^k, \to b =^i + 2 α^j - 2^k$ and $\to c = 2^i - α^j +^k$ and ${(\to a \times \to b) \times (\to b \times \to c)} \times (\to c \times \to a) = \to 0,$ then the value of $6 α$ is

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Let →a=α^i+2^j−3^k,→b=^i+2α^j−2^k and →c=2^i−α^j+^k and {(→a×→b)×(→b×→c)}×(→c×→a)=→0, then the value of 6α is

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