Geometric meaning of the quantity $|a|^2 |b|^2 |c|^2 - (a \cdot b)(b \cdot c)(c \cdot a)$ for non-coplanar vectors $a$, $b$, $c$ For two non-collinear vectors, $a$ and $b$, the quantity $$|a|^2 |b|^2 - (a \cdot b)(b \cdot a) = |a \times b|^2$$ is the square of the area of the parallelogram spanned by these two vectors. For three non-coplanar vectors, $a$, $b$ and $c$, we can form a similar expression $$B = |a|^2 |b|^2 |c|^2 - (a \cdot b) (b \cdot c) (c \cdot a)$$ which is not equal to the square of the volume of the parallelepiped spanned by these three vectors. 

Does $B$ have any geometrical (or other) meaning?

 A: I don't know if this hint is conclusive, but this is what I got:
\begin{align*}
B & = \lVert a\rVert^{2}\lVert b\rVert^{2}\lVert c\rVert^{2} - \langle a,b\rangle\langle b,c\rangle\langle c,a\rangle\\
& = \lVert a\rVert^{2}\lVert b\rVert^{2}\lVert c\rVert^{2} - \lVert a\rVert^{2}\lVert b\rVert^{2}\lVert c\rVert^{2}\cos(\theta_{1})\cos(\theta_{2})\cos(\theta_{3})\\
& = \lVert a\rVert^{2}\lVert b\rVert^{2}\lVert c\rVert^{2}[1 - \cos(\theta_{1})\cos(\theta_{2})\cos(\theta_{3})]
\end{align*}
A: I don't know a geometric interpretation for this quantity, but there's a natural interpretation of this quantity that makes it more transparently analogous to the other candidate you mentioned, namely the square of volume $V$ of the parallelepiped spanned by ${\bf a}, {\bf b}, {\bf c}$, at least for $n = 3$.
Recall that that squared area can be written as
$$\begin{align*}V^2 &= \operatorname{det}\pmatrix{{\bf a} \cdot {\bf a} & {\bf a} \cdot {\bf b} & {\bf a} \cdot {\bf c} \\ {\bf b} \cdot {\bf a} & {\bf b} \cdot {\bf b} & {\bf b} \cdot {\bf c} \\ {\bf c} \cdot {\bf a} & {\bf c} \cdot {\bf b} & {\bf c} \cdot {\bf c}} \\ &= ({\bf a} \cdot {\bf a})({\bf b} \cdot {\bf b})({\bf c} \cdot {\bf c}) - \mathfrak{S}[({\bf a} \cdot {\bf a})({\bf b} \cdot {\bf c})^2] + 2 ({\bf b} \cdot {\bf c}) ({\bf c} \cdot {\bf a})({\bf a} \cdot {\bf b}),\end{align*}$$ where $\mathfrak{S}[\cdot]$ denotes the cyclic sum of $\cdot$ in ${\bf a}, {\bf b}, {\bf c}$. 
On the other hand, recall that the determinant of an $n \times n$ matrix $M = (m_{ij})$ can be written as $$\det M = \sum_{\sigma} (\operatorname{sign} \sigma) m_{1 \sigma(1)} \cdots m_{n \sigma(n)},$$
where the sum is over all permutations $\sigma$ of $n$. Here, $\operatorname{sign} \sigma$ is $+1$ if the permutation is even and $-1$ if it is odd.
Now, if you know a little representation theory, you know $\operatorname{sign}$ is a representation of the group $S_n$ of permutations of $n$ elements. By replacing $\operatorname{sign}$ in the definition of the determinant with some other representation we get a generalization of the determinant (called an immanant) and---perhaps you see where this is going---can replace $\det$ in the first display equation above to get a new invariant for the (unordered) triple $({\bf a}, {\bf b}, {\bf c})$.
Taking $n = 3$ and the two-dimensional representation $\lambda$ gives the immanant
$$\operatorname{imm}_{\lambda}(M) = 2 m_{11} m_{12} m_{13} - m_{23} m_{31} m_{12} - m_{32} m_{13} m_{21} .$$ Replacing $\det$ in the first display equation with $\operatorname{imm}_{\lambda}$ gives exactly twice the quantity in question:
$$\boxed{\tfrac{1}{2} \operatorname{imm}_{\lambda} \pmatrix{{\bf a} \cdot {\bf a} & {\bf a} \cdot {\bf b} & {\bf a} \cdot {\bf c} \\ {\bf b} \cdot {\bf a} & {\bf b} \cdot {\bf b} & {\bf b} \cdot {\bf c} \\ {\bf c} \cdot {\bf a} & {\bf c} \cdot {\bf b} & {\bf c} \cdot {\bf c}} = ({\bf a} \cdot {\bf a})({\bf b} \cdot {\bf b})({\bf c} \cdot {\bf c}) - ({\bf b} \cdot {\bf c})({\bf c} \cdot {\bf a})({\bf a} \cdot {\bf b})} .$$
Remark Up to isomorphism there are only three irreducible representations of $S_3$, and the one remaining is the trivial representation. This gives rise to the permanent, $\operatorname{per}$, which is defined just by removing from the definition of $\det$ the sign of the permutation, so $$\begin{multline*}\operatorname{per}\pmatrix{{\bf a} \cdot {\bf a} & {\bf a} \cdot {\bf b} & {\bf a} \cdot {\bf c} \\ {\bf b} \cdot {\bf a} & {\bf b} \cdot {\bf b} & {\bf b} \cdot {\bf c} \\ {\bf c} \cdot {\bf a} & {\bf c} \cdot {\bf b} & {\bf c} \cdot {\bf c}} \\ = ({\bf a} \cdot {\bf a})({\bf b} \cdot {\bf b})({\bf c} \cdot {\bf c}) + \mathfrak{S}[({\bf a} \cdot {\bf a})({\bf b} \cdot {\bf c})^2] + 2 ({\bf b} \cdot {\bf c}) ({\bf c} \cdot {\bf a})({\bf a} \cdot {\bf b}),\end{multline*}$$ where $\mathfrak{S}[\cdot]$ denotes the cyclic sum of $\cdot$ in ${\bf a}, {\bf b}, {\bf c}$. It would be interesting to have a geometric interpretation of this quantity, too.
A: It occurred to me that if one seeks a geometrical interpretation of $B$, then one has to write it in terms of other geometrical quantities, i.e., length squared $|a|^2$, area squared $|a \times b|^2$ and volume squared $|a \cdot (b \times c)|^2$. Doing this, we find $$B = \frac{1}{2} \left[ |a|^2 |b \times c|^2 + |b|^2 |c \times a|^2 + |c|^2 |a \times b|^2 - |a \cdot (b \times c)|^2\right].$$ Now, $B$ vanishes only if all three vectors are collinear so it cannot represent any geometrical volume. On the other hand, the dimensionality of $B$ eliminates the possibility of describing some area. Therefore, I conclude that $B$ does not represent any simple geometrical quantity. 
