Vector Analysis Identity One Question given in class was to prove that:
$$\mathbf{(A \cdot B \times C)(a \cdot b \times c)} $$
is equal to
$$\begin{vmatrix} \mathbf A \cdot a & \mathbf A \cdot b & \mathbf A \cdot c \\
\mathbf B \cdot a & \mathbf B \cdot b & \mathbf B \cdot c \\ \mathbf C \cdot a & \mathbf C \cdot b & \mathbf C \cdot c \end{vmatrix}$$ 
I tried to write everything out in components, but that went no where fast. Anywhere to start? I am only allowed to use vector analysis and properties of the dot product and cross product. However, I do know that 
$$\mathbf{(A \cdot B \times C)} $$
is equal to
$$\begin{vmatrix} \mathbf A_x & \mathbf A_y & \mathbf A_z\\
\mathbf B_x & \mathbf B_y & \mathbf B_z\\ \mathbf C_x & \mathbf C_y & \mathbf C_z \ \end{vmatrix}$$ 
any help? It looks like it has some cool geometric properties. 

Just for reference in case someone want to see the actual algebraic manipulation.
In addition to some symmetry relations among dot and cross products
$$p \times q = -q \times p \quad\text{ and }\quad p\cdot(q \times r) = q \cdot( r \times p) = r \cdot( p \times q)$$
The key identities we need are:
$$\begin{align}
p \times (q \times r) &= (p\cdot r) q - (p \cdot q)r &\tag{*1}\\
(p \times q) \cdot ( r \times s) &= (p\cdot r)(q \cdot s) - (p\cdot s)(q \cdot r)& \tag{*2}
\end{align}$$
Using $(*2)$, we can rewrite the determinant at hand as
$$\left|\begin{matrix}
A\cdot a & A\cdot b & A\cdot c\\
B\cdot a & B\cdot b & B\cdot c\\
C\cdot a & C\cdot b & C\cdot c\\
\end{matrix}\right|
= \begin{array}{rl}
  & A\cdot a((B\cdot b)(C\cdot c)- (B\cdot c)(C\cdot b))\\
- & B\cdot a((A\cdot b)(C\cdot c)-(A\cdot c)(C\cdot b))\\
+ & C\cdot a((A\cdot b)(B\cdot c)-(A\cdot c)(B\cdot b))\end{array}
= \begin{array}{rl}
 & A\cdot a((B\times C)\cdot(b \times c))\\
-& B\cdot a((A\times C)\cdot(b \times c))\\
+& C\cdot a((A\times B)\cdot(b \times c))
\end{array}
$$
The determinant now has the form
$$u \cdot (b \times c)
\quad\text{ where }\quad
u = A\cdot a(B\times C)
- B\cdot a (A \times C)
+ C\cdot a (A \times B)$$
Using $(*1)$, we have
$$\begin{align}A\cdot a(B\times C) - B\cdot a(A\times C)
&= ((A\cdot a)B - (B\cdot a)A)\times C\\
&\stackrel{(*1)}{=} -(a \times (A \times B))\times C\\
&= C \times (a \times (A \times B))\\
&\stackrel{(*1)}{=} (C\cdot(A \times B)) a - (C\cdot a)(A\times B)
\end{align}
$$
Adding $(C\cdot a)(A \times B)$ to both sides, we obtain
$$u = (C\cdot (A \times B)) a = (A\cdot(B \times C)) a$$
and results follows.
