$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$  are 3 non-coplanar unit vectors inclined at an angle $\alpha (0<\alpha <\frac{\pi }{2})$  to each other. If the volume of tetrahedron formed by these vectors is $\frac{1}{\sqrt{360}}$  , then the value of  $\left|10(3{\cos }^2\alpha -2{\cos }^3\alpha )\right|$  

Given  $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=\cos \alpha$
$\therefore$   Volume of tetrahedron =  $\frac{1}{6}\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$
 Now,   ${\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]}^2=\left|\begin{array}{ccc}1& \cos \alpha & \cos \alpha \\ \cos \alpha & 1& \cos \alpha \\ \cos \alpha & \cos \alpha & 1\end{array}\right|$
  36 $V^2$=  $(1+2\cos \alpha ){(1-\cos \alpha )}^2$
$36\times \frac{1}{360}=(1+2\cos \alpha ){(1-\cos \alpha )}^2=\frac{1}{10}$  
 $\therefore 3{\cos }^2\alpha -2{\cos }^3\alpha =\frac{9}{10}$