Intersection of vectors to form a tetrahedron Suppose we have three unit vectors, namely $a,b,$ and $c,$ such that the angle between any of them is the acute angle $\theta.$ However, these vectors generate a tetrahedron with volume $\frac{1}{\sqrt{360}}.$ Find $$3\cos^2\theta - 2\cos^3\theta.$$

I've got little to no idea on how to start this problem due to the fact that it's a new concept to me. Can someone give me an overview of the basics and then a sketch of the problem for me to work on?
 A: Consider the following image

where the three angles in $V$ are $\theta$, develope the image in a plane, obtaining

Now we can observe that
\begin{align}
&\overline{AV}=\overline{BV}=\overline{CV}=1\\
&\overline{AB}=\overline{BC}=\overline{CA}=2\sin(\theta/2)\\
&\overline{HV}=\overline{KV}=\overline{LV}=\cos(\theta/2)\\
&\overline{AH}=\overline{BK}=\overline{CL}=\sqrt{3}\sin(\theta/2)\\
&\overline{H\Omega}=\overline{K\Omega}=\overline{L\Omega}=(\sqrt{3}/3)\sin(\theta/2)\\
\end{align}
then the height of the tetrahedron is
$$
h=\overline{V\Omega}=\sqrt{\overline{HV}^2-\overline{H\Omega}^2}=\sqrt{\cos^2(\theta/2)-\frac{1}{3}\sin^2(\theta/2)},
$$
while the base area is
$$
A_b=\frac{\sqrt{3}}{4}\overline{AB}^2=\sqrt{3}\sin^2(\theta/2).
$$
Finally, the volume is
$$
\mathscr{V}=\frac{1}{3}A_bh=\frac{\sqrt{3}}{3}\sin^2(\theta/2)\sqrt{\cos^2(\theta/2)-\frac{1}{3}\sin^2(\theta/2)}.
$$
Now observe that
\begin{align}
\mathscr{V}^2&=\frac{1}{3}\sin^4(\theta/2)\left(\cos^2(\theta/2)-\frac{1}{3}\sin^2(\theta/2)\right)=\\
&=\frac{1}{36}(1+2\cos\theta)(1-\cos\theta)^2=\\
&=\frac{1}{36}(1-3\cos^2\theta+2\cos^3\theta).
\end{align}
Now, from
$$
\mathscr{V}^2=\frac{1}{36}(1-3\cos^2\theta+2\cos^3\theta)=\frac{1}{360}
$$
we find
$$
3\cos^2\theta-2\cos^3\theta=\frac{9}{10}.
$$
