Prove that $\left(\overrightarrow{a} \times \overrightarrow{b}\right) \cdot \left(\overrightarrow{c} \times \overrightarrow{d}\right) + \left(\overrightarrow{b} \times \overrightarrow{c}\right) \cdot \left(\overrightarrow{a} \times \overrightarrow{d}\right) + \left(\overrightarrow{c} \times \overrightarrow{a}\right) \cdot \left(\overrightarrow{b} \times \overrightarrow{d}\right) = 0$
By definition,
$\begin{aligned}
\left(\overrightarrow{a} \times \overrightarrow{b}\right) \cdot \left(\overrightarrow{c} \times \overrightarrow{d}\right) & = \begin{vmatrix}
\overrightarrow{a} \cdot \overrightarrow{c} & \overrightarrow{a} \cdot \overrightarrow{d} \\\\
\overrightarrow{b} \cdot \overrightarrow{c} & \overrightarrow{b} \cdot \overrightarrow{d}
\end{vmatrix} \\\\
& = \left(\overrightarrow{a} \cdot \overrightarrow{c}\right) \left(\overrightarrow{b} \cdot \overrightarrow{d}\right) - \left(\overrightarrow{b} \cdot \overrightarrow{c}\right) \left(\overrightarrow{a} \cdot \overrightarrow{d}\right)
\end{aligned}$
$\begin{aligned}
\left(\overrightarrow{b} \times \overrightarrow{c}\right) \cdot \left(\overrightarrow{a} \times \overrightarrow{d}\right) & = \begin{vmatrix}
\overrightarrow{b} \cdot \overrightarrow{a} & \overrightarrow{b} \cdot \overrightarrow{d} \\\\
\overrightarrow{c} \cdot \overrightarrow{a} & \overrightarrow{c} \cdot \overrightarrow{d}
\end{vmatrix} \\\\
& = \left(\overrightarrow{b} \cdot \overrightarrow{a}\right) \left(\overrightarrow{c} \cdot \overrightarrow{d}\right) - \left(\overrightarrow{c} \cdot \overrightarrow{a}\right) \left(\overrightarrow{b} \cdot \overrightarrow{d}\right)
\end{aligned}$
$\begin{aligned}
\left(\overrightarrow{c} \times \overrightarrow{a}\right) \cdot \left(\overrightarrow{b} \times \overrightarrow{d}\right) & = \begin{vmatrix}
\overrightarrow{c} \cdot \overrightarrow{b} & \overrightarrow{c} \cdot \overrightarrow{d} \\\\
\overrightarrow{a} \cdot \overrightarrow{b} & \overrightarrow{a} \cdot \overrightarrow{d}
\end{vmatrix} \\\\
& = \left(\overrightarrow{c} \cdot \overrightarrow{b}\right) \left(\overrightarrow{a} \cdot \overrightarrow{d}\right) - \left(\overrightarrow{a} \cdot \overrightarrow{b}\right) \left(\overrightarrow{c} \cdot \overrightarrow{d}\right)
\end{aligned}$
$\therefore$ $\;$ $\left(\overrightarrow{a} \times \overrightarrow{b}\right) \cdot \left(\overrightarrow{c} \times \overrightarrow{d}\right) + \left(\overrightarrow{b} \times \overrightarrow{c}\right) \cdot \left(\overrightarrow{a} \times \overrightarrow{d}\right) + \left(\overrightarrow{c} \times \overrightarrow{a}\right) \cdot \left(\overrightarrow{b} \times \overrightarrow{d}\right) =$
$\;\;\;\;$ $\left(\overrightarrow{a} \cdot \overrightarrow{c}\right) \left(\overrightarrow{b} \cdot \overrightarrow{d}\right) - \left(\overrightarrow{b} \cdot \overrightarrow{c}\right) \left(\overrightarrow{a} \cdot \overrightarrow{d}\right) + \left(\overrightarrow{b} \cdot \overrightarrow{a}\right) \left(\overrightarrow{c} \cdot \overrightarrow{d}\right)$
$\;\;\;\;\;\;\;$ $- \left(\overrightarrow{c} \cdot \overrightarrow{a}\right) \left(\overrightarrow{b} \cdot \overrightarrow{d}\right) + \left(\overrightarrow{c} \cdot \overrightarrow{b}\right) \left(\overrightarrow{a} \cdot \overrightarrow{d}\right) - \left(\overrightarrow{a} \cdot \overrightarrow{b}\right) \left(\overrightarrow{c} \cdot \overrightarrow{d}\right)$
$= \left\{\left(\overrightarrow{a} \cdot \overrightarrow{c}\right) - \left(\overrightarrow{c} \cdot \overrightarrow{a}\right)\right\} \left(\overrightarrow{b} \cdot \overrightarrow{d}\right) + \left\{\left(\overrightarrow{c} \cdot \overrightarrow{b}\right) - \left(\overrightarrow{b} \cdot \overrightarrow{c}\right)\right\} \left(\overrightarrow{a} \cdot \overrightarrow{d}\right)$
$\;\;\;\;\;\;\;\;\;\;\;\;\;$ $+ \left\{\left(\overrightarrow{b} \cdot \overrightarrow{a}\right) - \left(\overrightarrow{a} \cdot \overrightarrow{b}\right)\right\} \left(\overrightarrow{c} \cdot \overrightarrow{d}\right)$
$= 0$ $\;\;\;$ $\because$ $\;$ $\overrightarrow{a} \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{a}$; $\;\;$ $\overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{c} \cdot \overrightarrow{b}$; $\;\;$ $\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{a}$
Hence proved.