Simplifying $|\vec{A} \times (\vec{B} \times \vec{C})|^2$ I know the vector identity $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C}) \vec{B} - (\vec{A} \cdot \vec{B}) \vec{C}$
Now, is there a succinct way of obtaining $|\vec{A} \times (\vec{B} \times \vec{C})|^2$ using vector algebra? I know we can expand, multiply and group the terms back, but is there a neater way of obtaining the result?
Using Levi-Civita symbols, would this be easier?
 A: Use triple product square equation or exterior algebra inner product to get
$$ (\vec{A} \cdot (\vec{B} \times \vec{C}))^2 = \begin{vmatrix} \vec{A}\cdot\vec{A} & \vec{A}\cdot\vec{B} & \vec{A}\cdot\vec{C} \\
\vec{B}\cdot\vec{A} & \vec{B}\cdot\vec{B} & \vec{B}\cdot\vec{C} \\
\vec{C}\cdot\vec{A} & \vec{C}\cdot\vec{B} & \vec{C}\cdot\vec{C} 
 \end{vmatrix}.\tag{1} $$
Now we also have the equation
$$ |\vec{A} \times (\vec{B} \times \vec{C})|^2 =
|\vec{A}|^2\  | \vec{B} \times \vec{C} |^2 -
(\vec{A} \cdot (\vec{B} \times \vec{C}))^2
\tag{2} $$
and a little bit of inspection leads to the result
$$ |\vec{A} \times (\vec{B} \times \vec{C})|^2 = - \begin{vmatrix} 0 & \vec{A}\cdot\vec{B} & \vec{A}\cdot\vec{C} \\
\vec{B}\cdot\vec{A} & \vec{B}\cdot\vec{B} & \vec{B}\cdot\vec{C} \\
\vec{C}\cdot\vec{A} & \vec{C}\cdot\vec{B} & \vec{C}\cdot\vec{C} 
 \end{vmatrix}.\tag{3} $$
The advantage of exterior algebra is that it works in $\ \mathbb{R}^n\ $ while the usual cross product is restricted to $\ \mathbb{R}^3.$
