Verify that $\left(\overrightarrow{a} \times \overrightarrow{b}\right) \times \left(\overrightarrow{c} \times \overrightarrow{d}\right) = \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{d}\right] \overrightarrow{c} - \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\right] \overrightarrow{d}$ if $\overrightarrow{a} = \hat{i} + \hat{j} + \hat{k}$, $\overrightarrow{b} = 2 \hat{i} + \hat{k}$, $\overrightarrow{c} = 2 \hat{i} + \hat{j} + \hat{k}$ and $\overrightarrow{d} = \hat{i} + \hat{j} + 2 \hat{k}$
$\begin{aligned}
\overrightarrow{a} \times \overrightarrow{b} & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 1 \\
2 & 0 & 1
\end{vmatrix} \\\\
& = \hat{i} \left(1 - 0\right) - \hat{j} \left(1 - 2\right) + \hat{k} \left(0 - 2\right) \\\\
& = \hat{i} + \hat{j} - 2 \hat{k}
\end{aligned}$
$\begin{aligned}
\overrightarrow{c} \times \overrightarrow{d} & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & 1 \\
1 & 1 & 2
\end{vmatrix} \\\\
& = \hat{i} \left(2 - 1\right) - \hat{j} \left(4 - 1\right) + \hat{k} \left(2 - 1\right) \\\\
& = \hat{i} - 3 \hat{j} + \hat{k}
\end{aligned}$
$\begin{aligned}
\therefore \; \left(\overrightarrow{a} \times \overrightarrow{b}\right) \times \left(\overrightarrow{c} \times \overrightarrow{d}\right) & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & -2 \\
1 & -3 & 1
\end{vmatrix} \\\\
& = \hat{i} \left(1 - 6\right) - \hat{j} \left(1 + 2\right) + \hat{k} \left(-3 - 1\right) \\\\
& = - 5 \hat{i} - 3 \hat{j} - 4 \hat{k} \;\;\; \cdots \; (1)
\end{aligned}$
$\left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{d}\right] = \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{d}\right)$
$\begin{aligned}
\overrightarrow{b} \times \overrightarrow{d} & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 0 & 1 \\
1 & 1 & 2
\end{vmatrix} \\\\
& = \hat{i} \left(0 - 1\right) - \hat{j} \left(4 - 1\right) + \hat{k} \left(2 - 0\right) \\\\
& = - \hat{i} - 3 \hat{j} + 2 \hat{k}
\end{aligned}$
$\begin{aligned}
\therefore \; \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{d}\right) & = \left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(- \hat{i} - 3 \hat{j} + 2 \hat{k}\right) \\\\
& = -1 - 3 + 2 \\\\
& = - 2
\end{aligned}$
$\begin{aligned}
\therefore \; \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{d}\right] \overrightarrow{c} & = -2 \left(2 \hat{i} + \hat{j} + \hat{k}\right) \\\\
& = - 4 \hat{i} - 2 \hat{j} - 2 \hat{k} \;\;\; \cdots \; (2a)
\end{aligned}$
$\left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\right] = \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{c}\right)$
$\begin{aligned}
\overrightarrow{b} \times \overrightarrow{c} & = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 0 & 1 \\
2 & 1 & 1
\end{vmatrix} \\\\
& = \hat{i} \left(0 - 1\right) - \hat{j} \left(2 - 2\right) + \hat{k} \left(2 - 0\right) \\\\
& = - \hat{i} + 2 \hat{k}
\end{aligned}$
$\begin{aligned}
\therefore \; \overrightarrow{a} \cdot \left(\overrightarrow{b} \times \overrightarrow{c}\right) & = \left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(- \hat{i} + 2 \hat{k}\right) \\\\
& = -1 + 2 \\\\
& = 1
\end{aligned}$
$\begin{aligned}
\therefore \; \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\right] \overrightarrow{d} & = 1 \left(\hat{i} + \hat{j} + 2 \hat{k}\right) \\\\
& = \hat{i} + \hat{j} + 2 \hat{k} \;\;\; \cdots \; (2b)
\end{aligned}$
$\therefore$ $\;$ We have from equations $(2a)$ and $(2b)$,
$\begin{aligned}
\left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{d}\right] \overrightarrow{c} - \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\right] \overrightarrow{d} & = \left(-4 \hat{i} - 2 \hat{j} - 2 \hat{k}\right) - \left(\hat{i} + \hat{j} + 2 \hat{k}\right) \\\\
& = - 5 \hat{i} - 3 \hat{j} - 4 \hat{k} \;\;\; \cdots \; (3)
\end{aligned}$
$\therefore$ $\;$ We have from equations $(1)$ and $(3)$,
$\left(\overrightarrow{a} \times \overrightarrow{b}\right) \times \left(\overrightarrow{c} \times \overrightarrow{d}\right) = \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{d}\right] \overrightarrow{c} - \left[\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\right] \overrightarrow{d}$
Hence proved.