Let a→,b→,c→ be three non-coplanar vectors and d→ be a non-zero vector which is perpendicular to a→+b→+c→. If d→=xb→×c→+yc→×a→+za→×b→, then

0=(a→+b→+c→)⋅d→=(a→+b→+c→)⋅(xb→×c→+yc→×a→+za→×b→)=(x+y+z)[a→b→c→]But [a→b→c→]≠0∴ x+y+z=0⇒x3+y3+z3=3xyz.

Let a→,b→,c→ be three non-coplanar vectors and d→ be a non-zero vector which is perpendicular to a→+b→+c→. If d→=xb→×c→+yc→×a→+za→×b→, then

0=(a→+b→+c→)⋅d→=(a→+b→+c→)⋅(xb→×c→+yc→×a→+za→×b→)=(x+y+z)[a→b→c→]But [a→b→c→]≠0∴ x+y+z=0⇒x3+y3+z3=3xyz.

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Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors and $\vec{d}$ be a non-zero vector which is perpendicular to $\vec{a} + \vec{b} + \vec{c}$ . If $\vec{d} = x \vec{b} \times \vec{c} + y \vec{c} \times \vec{a} + z \vec{a} \times \vec{b}$ , then

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$x^{3} + y^{3} + z^{3} = 3 x y z$

$x y + y z + z x = 0$

$x + y + z = 1$

$x = y = z$

answer is C.

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$0 = (\vec{a} + \vec{b} + \vec{c}) \cdot \vec{d}$

$\begin{array}{l} = (\vec{a} + \vec{b} + \vec{c}) \cdot (x \vec{b} \times \vec{c} + y \vec{c} \times \vec{a} + z \vec{a} \times \vec{b}) \\ = (x + y + z) [\vec{a} \vec{b} \vec{c}] \end{array}$

But $[\vec{a} \vec{b} \vec{c}] \neq 0$

$∴ x + y + z = 0 \Rightarrow x^{3} + y^{3} + z^{3} = 3 x y z$ .

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Let a→,b→,c→ be three non-coplanar vectors and d→ be a non-zero vector which is perpendicular to a→+b→+c→. If d→=xb→×c→+yc→×a→+za→×b→, then

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Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors and $\vec{d}$ be a non-zero vector which is perpendicular to $\vec{a} + \vec{b} + \vec{c}$ . If $\vec{d} = x \vec{b} \times \vec{c} + y \vec{c} \times \vec{a} + z \vec{a} \times \vec{b}$ , then