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Q.
Let $\vec{u}=\hat{i}-\hat{j}-2 \hat{k}, \vec{v}=2 \hat{i}+\hat{j}-\hat{k}, \vec{v} \cdot \vec{w}=2$ and $\vec{v} \times \vec{w}=\vec{u}+\lambda \vec{v}$. Then $\vec{u} \cdot \vec{w}$ is equal to
$\overrightarrow{ u }=(1,-1,-2), \overrightarrow{ v }=(2,1,-1), \overrightarrow{ v } \cdot \overrightarrow{ w }=2$
$ \overrightarrow{ v } \times \overrightarrow{ w }=\overrightarrow{ u }+\lambda \overrightarrow{ v } \ldots \ldots \ldots \ldots \ldots \ldots \ldots(1) $
Taking dot with $ \overrightarrow{ w } \text { in }(1) $
$ \overrightarrow{ w } \cdot(\overrightarrow{ v } \times \overrightarrow{ w })=\overrightarrow{ u } \cdot \overrightarrow{ w }+\lambda \overrightarrow{ v } \cdot \overrightarrow{ w }$
$ \Rightarrow 0=\overrightarrow{ u } \cdot \overrightarrow{ w }+2 \lambda$
Taking dot with $\vec{v}$ in (1)
$ \overrightarrow{ v } \cdot(\overrightarrow{ v } \times \overrightarrow{ w })=\overrightarrow{ u } \cdot \overrightarrow{ v }+\lambda \overrightarrow{ v } \cdot \overrightarrow{ v }$
$ \Rightarrow 0=(2-1+2)+\lambda \cdot(6)$
$ \lambda=-\frac{1}{2} $
$ \Rightarrow \overrightarrow{ u } \cdot \overrightarrow{ w }=-2 \lambda=1$