Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such
that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$. if
$\lambda=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ and
$\vec{d}=\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}$, then
the ordered pair, $(\lambda, \vec{d})$ is equal to:
Correct Option: , 4
$|\vec{a}+\vec{b}+\vec{c}|^{2}=0$
$3+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0$
$(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=\frac{-3}{2} \Rightarrow \lambda=\frac{-3}{2}$
$\vec{d}=\vec{a} \times \vec{b}+\vec{b} \times(-\vec{a}-\vec{b})+(-\vec{a}-\vec{b}) \times \vec{a}[\because \vec{c}=-\vec{a}-\vec{b}]$
$=\vec{a} \times \vec{b}+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}$
$\vec{d}=3(\vec{a} \times \vec{b})$
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