$\displaystyle \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\displaystyle \overrightarrow{\mathrm{c}}$ are three mutually perpendicular unit vectors. If $\displaystyle \theta$ is the angle between $\displaystyle \overrightarrow{\mathrm{a}}$ and ( $\displaystyle 2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{~b}}+6 \overrightarrow{\mathrm{c}}$ ), find the value of $\displaystyle \cos \theta$.

$$\begin{aligned} & \text { Given }|\vec{\mathrm{a}}|=|\vec{\mathrm{b}}|=|\vec{\mathrm{c}}|=1 \text { and } \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=\vec{\mathrm{b}} \cdot \vec{\mathrm{c}}=\vec{\mathrm{c}} \cdot \vec{\mathrm{a}}=0 \\ & \operatorname{Now},|2 \vec{\mathrm{a}}+3 \vec{\mathrm{b}}+6 \vec{\mathrm{c}}|^{2}=4|\vec{\mathrm{a}}|^{2}+9|\vec{\mathrm{b}}|^{2}+36|\vec{\mathrm{c}}|^{2}=49 \\ & \Rightarrow|2 \vec{\mathrm{a}}+3 \vec{\mathrm{b}}+6 \vec{\mathrm{c}}|=7 \\ & \cos \theta=\frac{\vec{\mathrm{a}} \cdot(2 \vec{\mathrm{a}}+3 \vec{\mathrm{b}}+6 \vec{\mathrm{c}})}{|\vec{\mathrm{a}}||2 \vec{\mathrm{a}}+3 \vec{\mathrm{b}}+6 \vec{\mathrm{c}}|}=\frac{2|\vec{\mathrm{a}}|^{2}}{|\vec{\mathrm{a}}||2 \vec{\mathrm{a}}+3 \vec{\mathrm{b}}+6 \vec{\mathrm{c}}|} \\ & \therefore \cos \theta=\frac{2}{7} \end{aligned} $$