Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $\theta$, with the vector $\vec{a}+\vec{b}+\vec{c}$. Then $36 \cos ^{2} 2 \theta$ is equal to
$|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^{2}=|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{c}}|^{2}+2(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}})$
$=3$
$\Rightarrow|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3}$
$\vec{a} \cdot(\vec{a}+\vec{b}+\vec{c})=|\vec{a}|+|\vec{a}+\vec{b}+\vec{c}| \cos \theta$
$\Rightarrow 1=\sqrt{3} \cos \theta$
$\Rightarrow \cos 2 \theta=-\frac{1}{3}$
$\Rightarrow 36 \cos ^{2} 2 \theta=4$