Given vectors are:
$\begin{aligned} & \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-3 \overline{\mathrm{j}}-\hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{b}}=\overline{\mathrm{i}}+4 \overline{\mathrm{j}}-2 \overrightarrow{\mathrm{k}} \\ \Rightarrow \quad & \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\hat{\mathrm{k}})+(\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \\ &=3 \overline{\mathrm{i}}+\overline{\mathrm{j}}-3 \hat{\mathrm{k}} \end{aligned}$
and $\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}=(2 \hat{\mathrm{i}}-3 \overline{\mathrm{j}}-\hat{\mathrm{k}})-(\overline{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})$
$=\overline{\mathrm{i}}-7 \overline{\mathrm{j}}+\hat{\mathrm{k}}$
$\therefore(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}})=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 1 & -3 \\ 1 & -7 & 1\end{array}\right|$
$=\hat{\mathrm{i}}\left|\begin{array}{cc}1 & -3 \\ -7 & 1\end{array}\right|-\hat{\mathrm{j}}\left|\begin{array}{cc}3 & -3 \\ 1 & 1\end{array}\right|+\widehat{k}\left|\begin{array}{cc}3 & 1 \\ 1 & -7\end{array}\right|$
$=\hat{\mathrm{i}}(1-21)-\hat{\mathrm{j}}(3+3)+\mathrm{k}(-21-1)$
$=-20 \hat{\mathrm{i}}-6 \widehat{\mathrm{j}}-22 \hat{\mathrm{k}}$
$=-2(10 \hat{\mathrm{i}}+3 \widehat{\mathrm{j}}+11 \widehat{\mathrm{k}})$
Now, $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 4 & -2\end{array}\right|$
$=\hat{i}\left|\begin{array}{cc}-3 & -1 \\ 4 & -2\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & -1 \\ 1 & -2\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & -3 \\ 1 & 4\end{array}\right|$
$=\widehat{\mathrm{i}}(6+4)-\hat{\mathrm{j}}(-4+1)+\mathrm{k}(8+3)$
$=10 \hat{\mathrm{i}}+3 \widehat{\mathrm{j}}+11 \hat{\mathrm{k}}$
Hence, $(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=-2(\vec{a} \times \vec{b})$