Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$. If the vector $\vec{C}$ is perpendicular to $\vec{b}$ and $\vec{a} \cdot \vec{c}=5$, then $|\vec{c}|$ is equal to

<p>$$\begin{aligned} & \overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{b}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})) \\ & =\lambda((\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{~b}}) \overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{~b}}) \overrightarrow{\mathrm{b}}) \\ & =\lambda(11 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{~b}})=\lambda(11 \mathrm{i}+22 \mathrm{j}+33 \mathrm{k}-6 \mathrm{i}-2 \mathrm{j}+2 \mathrm{k}) \\ & =\lambda(5 \mathrm{i}+20 \mathrm{j}+35 \mathrm{k}) \\ & =5 \lambda(5 \mathrm{i}+4 \mathrm{j}+7 \mathrm{k}) \\ & =\text { Given } \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=5 \\ & =5 \lambda(1+8+21)=5=\lambda=\frac{1}{30} \\ & \Rightarrow \overrightarrow{\mathrm{c}}=\frac{1}{6}(\mathrm{i}+4 \mathrm{j}+7 \mathrm{k}) \\ & |\overrightarrow{\mathrm{c}}|=\frac{\sqrt{1+16+49}}{6}=\sqrt{\frac{11}{6}} \end{aligned}$$</p>