Let $$\overrightarrow u = \widehat i - \widehat j - 2\widehat k,\overrightarrow v = 2\widehat i + \widehat j - \widehat k,\overrightarrow v .\,\overrightarrow w = 2$$ and $$\overrightarrow v \times \overrightarrow w = \overrightarrow u + \lambda \overrightarrow v $$. Then $\overrightarrow u .\,\overrightarrow w$ is equal to :

$$ \begin{aligned} &\begin{aligned} & \vec{v} \times \vec{w}=(\vec{u}+\lambda \vec{v})=\hat{i}-\hat{j}-2 \hat{k}+\lambda(2 \hat{i}+\hat{j}-\hat{k}) \\\\ & =(2 \lambda+1) \hat{i}+(\lambda-1) \hat{j}-(2+\lambda) \hat{k} \\ & \end{aligned}\\ &\begin{aligned} & \text { Now, } \vec{v} \cdot(\vec{v} \times \vec{w})=0 \\\\ & \Rightarrow(2 \hat{i}+\hat{j}-\hat{k}) \cdot((2 \lambda+1) \hat{i}+(\lambda-1) \hat{j}-(\lambda+2) \hat{k})=0 \\\\ & \Rightarrow2(2 \lambda+1)+\lambda-1+\lambda+2=0 \Rightarrow 6 \lambda+3=0 \\\\ & \Rightarrow \lambda=-\frac{1}{2} \\\\ & \vec{w} \cdot(\vec{v} \times \vec{w})=\vec{w} \cdot(\vec{u}+\lambda \vec{v})=\vec{u} \cdot \vec{w}+\lambda \vec{v} \cdot \vec{w}=0 \\\\ & \vec{u} \cdot \vec{w}=-\lambda \vec{v} \cdot \vec{w} \\\\ & \vec{u} \cdot \vec{w}=\frac{1}{2} \times 2=1 \end{aligned} \end{aligned} $$