How to prove that a sum of quintuple cross products is equal to zero? Show that : 
$$
p \times [(a \times q) \times (b \times r)] \\
+ q \times [(a \times r) \times (b \times p)] \\
+ r \times [(a \times p) \times (b \times q)] = 0
$$
where $\times$ is cross product and $a, b, p, q,$ and $r$ are vectors.
 A: We have to prove that :
$$\underbrace{\left(\begin{array}{r}P \times [(\color{red}{A} \times Q) \times (\color{red}{B} \times R)] \\
+ \ Q \times [(\color{red}{A} \times R) \times (\color{red}{B} \times P)] \\
+ \ R \times [(\color{red}{A} \times P) \times (\color{red}{B} \times Q)]\end{array}\right)}_{L}= 0\tag{1}$$
I will prove the result in the case where $P,Q,R$ are independent, i.e., constitute a basis of $\mathbb{R^3}$.
I will use the following notations : $|ABC|$ for the determinant with columns $A,B,C$ in that order and 


*

*$\cdot$ for the dot product,

*'*' for multiplication of a vector by a number,

*and the ordinary dot for multiplication of two numbers.
Let us recall the rather classical vector quadruple relationship :
$$(A \times B) \times (C \times D)=|ABD|*C - |ABC|*D\tag{2}$$
(that can be as well found on pages 76-77 of the very clear 1901 book by Gibbs/Wilson).
Let us use (2) three times so as to transform the LHS $L$ of (1) into :
$$L=\left(\begin{array}{r}P \times (|AQR|*B-|AQB|*R)\\+ \ Q \times (|ARP|*B-|ARB|*P)\\+ \ R \times (|APQ|*B-|APB|*Q)\end{array}\right)$$
We have to show that $L$ is zero.
With this purpose in mind, as we have assumed that $\{P,Q,R\}$ constitute a basis of $\mathbb{R^3}$, it suffices to show that the three dot products $L \cdot P, L \cdot Q, L \cdot R$  are $0$.
Let us show that $L \cdot P=0.$
We recall that $(U \times V) \cdot W=|UVW|$. We can write 
$$L \cdot P=0 \ + \ |QBP|.|ARP| \ + \ 0 \ + \ |RBP|.|APQ| \ -\ |RQP|.|APB|$$
Why is this expression $0$ ? Because of Grassmann-Plücker relationships as given for example in Is Grassmann-Plucker relation implied by 3-term Grassmann-Plucker relation? with $r=3$ :
$$|x_1x_2x_3|.|y_1y_2x_3|-|x_1y_1x_3|.|x_2y_2x_3|+|x_1y_2x_3|.|y_1x_2x_3|=0$$
(take $y_1=A, y_2=B, x_3=P, x_2=R, x_1=Q$ and use properties of determinants).
I will prove neither $L.Q=0$, nor $L.R=0$ because the same technique is applicable.
Remarks : 
1) Relationship (1) is identical to Tomihisa identity given in page 30 of this very nice article by S. Tabachnikov. I will try to establish the link with our problem.
2) Another mention of the Grassmann-Pluecker relationships can be found in the following text that I reproduce as an image (Muir, Theory of Determinants, Vol 5, p. 5) :

