If a¯=2i¯+j¯+k¯,b¯=i¯+2j¯+2k¯, c¯=i¯+j¯+2k¯ and a¯×(b¯×c¯)= (1+α)i¯+β(1+α)j¯+γ(1+α)(1+β)k¯ then

a¯×(b¯×c¯)=−i¯+4j¯−2k¯⇒1+α=−1,β(1+α)=4,γ(1+α)(1+β)=−2α=−2,β(−1)=4⇒β=−4⇒γ(−1)(1−4)=−2⇒γ(−1)(−3)=−2⇒γ=−23

If a¯=2i¯+j¯+k¯,b¯=i¯+2j¯+2k¯, c¯=i¯+j¯+2k¯ and a¯×(b¯×c¯)= (1+α)i¯+β(1+α)j¯+γ(1+α)(1+β)k¯ then

a¯×(b¯×c¯)=−i¯+4j¯−2k¯⇒1+α=−1,β(1+α)=4,γ(1+α)(1+β)=−2α=−2,β(−1)=4⇒β=−4⇒γ(−1)(1−4)=−2⇒γ(−1)(−3)=−2⇒γ=−23

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If $\bar{a} = 2 \bar{i} + \bar{j} + \bar{k}, \bar{b} = \bar{i} + 2 \bar{j} + 2 \bar{k}$ , $\bar{c} = \bar{i} + \bar{j} + 2 \bar{k}$ and $\bar{a} \times (\bar{b} \times \bar{c}) =$ $(1 + α) \bar{i} + β (1 + α) \bar{j} + γ (1 + α) (1 + β) \bar{k}$ then

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$α = - 2, β = - 4, γ = \frac{- 2}{3}$

$α = 2, β = 4, γ = \frac{- 2}{3}$

$α = 2, β = - 4, γ = \frac{2}{3}$

$α = - 2, β = 4, γ = \frac{2}{3}$

answer is A.

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Detailed Solution

$\bar{a} \times (\bar{b} \times \bar{c}) = - \bar{i} + 4 \bar{j} - 2 \bar{k}$

$\Rightarrow 1 + α = - 1, β (1 + α) = 4, γ (1 + α) (1 + β) = - 2$

$α = - 2, β (- 1) = 4 \Rightarrow β = - 4$

$\Rightarrow γ (- 1) (1 - 4) = - 2$

$\Rightarrow γ (- 1) (- 3) = - 2 \Rightarrow γ = \frac{- 2}{3}$

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If a¯=2i¯+j¯+k¯,b¯=i¯+2j¯+2k¯, c¯=i¯+j¯+2k¯ and a¯×(b¯×c¯)= (1+α)i¯+β(1+α)j¯+γ(1+α)(1+β)k¯ then

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If $\bar{a} = 2 \bar{i} + \bar{j} + \bar{k}, \bar{b} = \bar{i} + 2 \bar{j} + 2 \bar{k}$ , $\bar{c} = \bar{i} + \bar{j} + 2 \bar{k}$ and $\bar{a} \times (\bar{b} \times \bar{c}) =$ $(1 + α) \bar{i} + β (1 + α) \bar{j} + γ (1 + α) (1 + β) \bar{k}$ then