Let \(\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}+\hat{j}\) and \(\vec{c}\) be a vector such that \(|\vec{c}-\vec{a}|=3\). If \(\overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\), then the angle between \(\overrightarrow{\mathrm{p}}\) and \(\overrightarrow{\mathrm{c}}\) is \(\frac{\pi}{6}\) and \(|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{c}}|=3\). Thus \(\vec{a} . \vec{c}\) is equal to-

Given \(\vec{a}=2 \hat{i}+\hat{j}-2 \widehat{k}, \vec{b}=\hat{i}+\hat{j},|\vec{c}-\vec{a}|=3\)
\(\vec{p}=\vec{a} \times \vec{b},|\vec{p} \times \vec{c}|=3\) and angle between \(\vec{p}\) and \(\vec{c}\) is \(\frac{\pi}{6}\)
Here, \(\vec{p}=\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \widehat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{i}-2 \hat{j}+\widehat{k} \Rightarrow|\vec{p}|=3\)
also \(|\vec{a}|=3\) and \(|\vec{b}|=\sqrt{2}\)
\(\begin{aligned} & \text { Now } \because|\vec{p} \times \vec{c}|=3 \Rightarrow|\vec{p}||\vec{c}| \sin \frac{\pi}{6}=3 \\ & \Rightarrow 3|\vec{c}| \times \frac{1}{2}=3 \\ & \Rightarrow|\vec{c}|=2 \\ & \text { and }|\vec{c}-\vec{a}|=3 \Rightarrow|\vec{c}-\vec{a}|^2=9 \\ & \Rightarrow|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a}=9 \\ & \Rightarrow 4+9-2 \vec{a} \cdot \vec{c}=9\end{aligned}\)
\(\Rightarrow \vec{a} \cdot \vec{c}=2\)