Let \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\) be two vectors. If \(\vec{c}\) is a vector such that \(\vec{b} \times \vec{c}=\vec{b} \times \vec{a}\) and \(\vec{c} \cdot \vec{a}\) then \(\vec{c} \cdot \vec{b}\) is equal to

\(\begin{aligned} & \vec{b} \times \vec{c}=\vec{b} \times \vec{a} \\ & \Rightarrow \vec{b} \times \vec{c}-\vec{b} \times \vec{a}=\overrightarrow{0} \\ & \Rightarrow \vec{b} \times(\vec{c}-\vec{a})=\overrightarrow{0} \\ & \Rightarrow \vec{c}-\vec{a} \| \vec{b} \\ & \Rightarrow \vec{c}-\vec{a}=\lambda \vec{b} \\ & \Rightarrow \vec{c}=\vec{a}+\lambda \vec{b} \\ & \because \vec{c} \cdot \vec{a}=0 \Rightarrow(\vec{a}+\lambda \vec{b}) \cdot \vec{a}=0 \Rightarrow \vec{a} \cdot \vec{b}+\lambda \vec{b} \cdot \vec{a}=0 \\ & \Rightarrow 11^2+(-2)^2+1^2+\lambda(1 \times 1+(-2) \times(-1)+1 \times 1) \\ & \Rightarrow \lambda=-\frac{3}{2}\end{aligned}\)
\(\begin{aligned} & \text { Now } \vec{c} \cdot \vec{b}=(\vec{a}+\lambda \vec{b}) \cdot \vec{b} \\ & =\vec{a} \cdot \vec{b}+\lambda \vec{b} \cdot \vec{b} \\ & =4-\frac{3}{2} \times 3=4-\frac{9}{2}=-\frac{1}{2}\end{aligned}\)