For non zero vectors a→,b→,c→, |(a→×b→)⋅c→|=|a→||b→||c→| holds if and only if

If θ is the angle between a→ and b→and ϕ is the angle between a→×b→and c→, then(a→×b→)⋅c→=|a→||b→||c→|sin⁡θcos⁡ϕ∴ sin⁡θ=1,cos⁡ϕ=1⇒θ=π2,ϕ=0∴ a→⊥b→,c→ is along a→×b→∴ c→⊥a→,c→⊥b→.a→⋅b→=b→⋅c→=c→⋅a→=0

For non zero vectors a→,b→,c→, |(a→×b→)⋅c→|=|a→||b→||c→| holds if and only if

If θ is the angle between a→ and b→and ϕ is the angle between a→×b→and c→, then(a→×b→)⋅c→=|a→||b→||c→|sin⁡θcos⁡ϕ∴ sin⁡θ=1,cos⁡ϕ=1⇒θ=π2,ϕ=0∴ a→⊥b→,c→ is along a→×b→∴ c→⊥a→,c→⊥b→.a→⋅b→=b→⋅c→=c→⋅a→=0

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For non zero vectors $\vec{a}, \vec{b}, \vec{c},$ $| (\vec{a} \times \vec{b}) \cdot \vec{c} | = | \vec{a} | | \vec{b} | | \vec{c} |$ holds if and only if

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$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = 0$

$\vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$

$\vec{c} \cdot \vec{a} = \vec{a} \cdot \vec{b} = 0$

$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$

answer is D.

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Detailed Solution

If $θ$ is the angle between $\vec{a}$ and $\vec{b}$ and $ϕ$ is the angle between $\vec{a} \times \vec{b}$ and ${\vec{c}}_{,}$ then
$(\vec{a} \times \vec{b}) \cdot \vec{c} = | \vec{a} | | \vec{b} | | \vec{c} | \sin θ \cos ϕ$
$∴ \sin θ = 1, \cos ϕ = 1 \Rightarrow θ = \frac{π}{2}, ϕ = 0$
$∴ \vec{a} ⊥ \vec{b}, \vec{c}$ is along $\vec{a} \times \vec{b}$
$∴ \vec{c} ⊥ \vec{a}, \vec{c} ⊥ \vec{b}$ .
$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0$