Given that,
$$
\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+3 \hat{\mathbf{k}}
$$
Now, $\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -1 \\ 1 & 0 & 3\end{array}\right|$
$$
\begin{aligned}
& =\hat{\mathbf{i}}(3-0)-\hat{\mathbf{j}}(6+1)+\hat{\mathbf{k}}(0-1) \\
& =3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}}
\end{aligned}
$$
$$
\begin{aligned}
& \text { Now, }\left[\begin{array}{lll}
\overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{b}} & \mathbf{c}
\end{array}\right]=\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}) \\
& =|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}| \cos \theta \\
& =1\left(\sqrt{3^2+7^2+1^2}\right) \cos \theta \\
& =\sqrt{59} \cos \theta \\
& \Rightarrow[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]_{\max }=\sqrt{59} \cdot 1 \\
&
\end{aligned}
$$
$(\because$ maximum value of $\cos \theta$ is 1 )
Hence, maximum value is $\sqrt{59}$.