$$
\begin{aligned}
& \text { Let } \mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\
& \qquad \begin{aligned}
|\mathbf{a}| & =\sqrt{2^2+1^2+(-2)^2}=3 \text { and } \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}} \\
|\mathbf{b}| & =\sqrt{1+1}=\sqrt{2} \\
\mathbf{a} \times \mathbf{b} & =\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
2 & 1 & -2 \\
1 & 1 & 0
\end{array}\right|=\hat{\mathbf{i}}(2)-\hat{\mathbf{j}}(2)+\hat{\mathbf{k}}(1)=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
|\mathbf{a} \times \mathbf{b}| & =\sqrt{2^2+(-2)^2+(1)^2}=3
\end{aligned}
\end{aligned}
$$
Now, $\mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|$ and $|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$
Squaring both side
$$
\begin{gathered}
|\mathbf{c}-\mathbf{a}|^2=(2 \sqrt{2})^2 \\
\Rightarrow \quad\left|\mathbf{c}^2\right|+\left|\mathbf{a}^2\right|-2 \mathbf{c} \cdot \mathbf{a}=8 \\
\Rightarrow|\mathbf{c}|^2+9-2|\mathbf{c}|=8 \Rightarrow|\mathbf{c}|^2-2|\mathbf{c}|+1=0
\end{gathered}
$$
Let
$$
|\mathbf{c}|=x
$$
$\begin{aligned} & \Rightarrow \quad x^2-2 x+1=0 \Rightarrow(x-1)^2=0 \\ & \Rightarrow \quad x=1 \Rightarrow|\mathbf{c}|=1 \\ & \text { Now, }|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \sin 30^{\circ} \\ & \\ & \end{aligned}$