We have,
$$
\begin{aligned}
& \mathbf{a}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\
& \mathbf{b}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\
& \mathbf{c}=\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}
\end{aligned}
$$
Now,
$$
\begin{aligned}
& (\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{b}-[\mathbf{a b} \mathbf{b}] \mathbf{c}=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{b} \\
& (\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a})=[\mathbf{b} \mathbf{c} \mathbf{a}] \mathbf{c}-[\mathbf{b} \mathbf{c}] \mathbf{a}=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{c} \\
& (\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})=[\mathbf{c} \mathbf{a} \mathbf{b}] \mathbf{a}-[\mathbf{c} \mathbf{a} \mathbf{a}] \mathbf{b}=[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{a} \\
& \text { Now, } \quad[\mathbf{a} \mathbf{b} \mathbf{c}]=\left[\begin{array}{ccc}
1 & -2 & -3 \\
2 & 1 & -1 \\
1 & 3 & -2
\end{array} \mid\right. \\
& =1(-2+3)+2(-4+1)-3(6-1)=1-6-15=-20
\end{aligned}
$$
$\begin{aligned} & \therefore[(\mathbf{a} \times \mathbf{b}) \times(\mathbf{b} \times \mathbf{c})(\mathbf{b} \times \mathbf{c}) \times(\mathbf{c} \times \mathbf{a}) \\ & =\left[\begin{array}{ll}-20 \mathbf{b}-20 \mathbf{c}-20 \mathbf{a}] \\ (\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})]\end{array}\right. \\ & =\left|\begin{array}{ccc}-40 & -20 & 20 \\ -20 & -60 & 40 \\ -20 & 40 & 60\end{array}\right| \\ & =-40[-3600-1600]+20[-1200+800] \\ & +20[-800-1200] \\ & =208000-8000-40000=160000 .\end{aligned}$