If a→·b→=1,b→·c→=2 and c→·a→=3, then the value of [a→×(b→×c→) b×(c→×a→) c→×(b→×a→)] is :

a→×(b→×c→)=(a→⋅c→)b→−(a→⋅b→)c→=3b→−c→b→×(c→×a→)=(b→⋅a→)c→−(b→⋅c→)a→=c→−2a→c→×(b→×a→)=(c→⋅a→)b→−(c→⋅b→)a→=3b→−2a→[3b→−c→,c→−2a→,3b→−2a→]=03-1-201-230a→ b→ c→=0

If a→·b→=1,b→·c→=2 and c→·a→=3, then the value of [a→×(b→×c→) b×(c→×a→) c→×(b→×a→)] is :

a→×(b→×c→)=(a→⋅c→)b→−(a→⋅b→)c→=3b→−c→b→×(c→×a→)=(b→⋅a→)c→−(b→⋅c→)a→=c→−2a→c→×(b→×a→)=(c→⋅a→)b→−(c→⋅b→)a→=3b→−2a→[3b→−c→,c→−2a→,3b→−2a→]=03-1-201-230a→ b→ c→=0

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If $\vec{a} \cdot \vec{b} = 1, \vec{b} \cdot \vec{c} = 2$ and $\vec{c} \cdot \vec{a} = 3$ , then the value of $[\vec{a} \times (\vec{b} \times \vec{c}) b \times (\vec{c} \times \vec{a}) \vec{c} \times (\vec{b} \times \vec{a})]$ is :

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$- 6 \vec{a} \cdot (\vec{b} \times \vec{c})$

$12 \vec{c} \cdot (\vec{a} \times \vec{b})$

$- 12 \vec{b} \cdot (\vec{c} \times \vec{a})$

answer is A.

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Detailed Solution

$\begin{array}{l} \vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = 3 \vec{b} - \vec{c} \\ \vec{b} \times (\vec{c} \times \vec{a}) = (\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a} = \vec{c} - 2 \vec{a} \\ \vec{c} \times (\vec{b} \times \vec{a}) = (\vec{c} \cdot \vec{a}) \vec{b} - (\vec{c} \cdot \vec{b}) \vec{a} = 3 \vec{b} - 2 \vec{a} \\ [3 \vec{b} - \vec{c}, \vec{c} - 2 \vec{a}, 3 \vec{b} - 2 \vec{a}] = |\begin{matrix} 0 & 3 & - 1 \\ - 2 & 0 & 1 \\ - 2 & 3 & 0 \end{matrix}| [\vec{a} \vec{b} \vec{c}] = 0 \end{array}$