Let $\overline{a},\overline{b},\overline{c}$ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle $\theta$ with the vector $\overline{a}+\overline{b}+\overline{c}$ then $36{\cos }^{2}2\theta =-----$

$\left|\overline{a}\right|=\left|\overline{b}\right|=\left|\overline{c}\right|=k,\overline{a}.\overline{b}=\overline{b}.\overline{c}=\overline{c}.\overline{a}=0$
${|\overline{a}+\overline{b}+\overline{c}|}^2=3{\left|\overline{a}\right|}^2.$
$|\overline{a}+\overline{b}+\overline{c}|=\sqrt{3}\left|\overline{a}\right|=\sqrt{3}k$
$\cos \theta =\frac{\overline{a}.(\overline{a}+\overline{b}+\overline{c})}{\left|\overline{a}\right||\overline{a}+\overline{b}+\overline{c}|}=\frac{{\left|\overline{a}\right|}^2+0+0}{{\left|\overline{a}\right|}^2.\sqrt{3}\left|\overline{a}\right|}=\frac{1}{\sqrt{3}}$
$36{\cos }^{2}2\theta =36{[2{\cos }^2\theta -1]}^2$
$=36{[\frac{2}{3}-1]}^2=36.\frac{1}{9}=4.$