Let $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=4 \hat{i}+\hat{j}+7 \hat{k}$ and $\vec{c}=\hat{i}-3 \hat{j}+4 \hat{k}$ be three vectors. If a vectors $\vec{p}$ satisfies $\vec{p} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{p} \cdot \vec{a}=0$, then $\vec{p} \cdot(\hat{i}-\hat{j}-\hat{k})$ is equal to

<p>$$\begin{aligned} & \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\ & (\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\ & \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}+\lambda \overrightarrow{\mathrm{b}} \end{aligned}$$</p> <p>Now, $\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{a}}=0$ (given)</p> <p>$$\begin{aligned} & \text { So, } \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0 \\ & (3-3-8)+\lambda(12+1-14)=0 \\ & \lambda=-8 \\ & \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}-8 \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{p}}=-31 \hat{\mathrm{i}}-11 \hat{\mathrm{j}}-52 \hat{\mathrm{k}} \\ & \text { So, } \overrightarrow{\mathrm{p}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & =-31+11+52 \\ & =32 \end{aligned}$$</p>