Since, \(\mathbf{A}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) and \(\mathbf{B}=\hat{\mathbf{i}}+\hat{\mathbf{j}}\) so
\(\begin{aligned}
& \mathbf{A} \times \mathbf{B}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
2 & 1 & -2 \\
1 & 1 & 0
\end{array}\right|=\hat{\mathbf{i}}(2)-\hat{\mathbf{j}}(\mathrm{Z})+\hat{\mathbf{k}}(2-1) \\
& =2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
& \because|(\mathbf{A} \times \mathbf{B}) \times \mathbf{C}|=|\mathbf{A} \times \mathbf{B}||\mathbf{C}| \sin 30^{\circ}
\end{aligned}\)
[as angle between \(\mathbf{A} \times \mathbf{B}\) and \(\mathbf{C}=30^{\circ}\) (given)]
\(\begin{aligned}
& =\sqrt{4+4+1}|\mathrm{C}|\left(\frac{1}{2}\right)=\frac{3}{2}|\mathrm{C}| \quad \ldots (i) \\
& \because \quad|\mathbf{C}-\mathbf{A}|=2 \sqrt{2} \quad \text{(Given)} \\
& \Rightarrow|\mathbf{C}|^2+|\mathbf{A}|^2-2 \mathbf{C} \cdot \mathbf{A}=8 \\
& \Rightarrow \quad|C|^2+9-2|C|=8 \quad \text{[as } \mathbf{A} \cdot \mathbf{C}=|\mathbf{C}|] \\
& \Rightarrow(|\mathbf{C}|-1)^2=0 \Rightarrow|\mathbf{C}|=1 \\
\end{aligned}\)
So, \(|(\mathbf{A} \times \mathbf{B}) \times \mathbf{C}|=\frac{3}{2}\) [from Eq. (i)]
Hence, option (2) is correct.