If \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}\) and \(\overrightarrow{\mathbf{c}}\) are non-zero coplanar vectors, then \([2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} 3 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}} 4 \overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}]\) is

\(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}\) non-zero coplanar vectors
\[
\begin{gathered}
{[2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} 3 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}} 4 \overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}]} \\
=(2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \cdot[(3 \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \times(4 \overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}})] \\
=(2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \cdot[12 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}-4 \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{c}}-3 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}]
\end{gathered}
\]
\(\begin{aligned}=(2 \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}})[12 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}-3 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}] \\ &(\because \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{c}}=0) \end{aligned}\)
\[
\begin{aligned}
&=24 \overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})-6 \overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}})+2 \overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}) \\
&\quad-12 \overrightarrow{\mathbf{b}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})+3 \overrightarrow{\mathbf{b}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}})-\overrightarrow{\mathbf{b}} \cdot(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}) \\
&=24[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]-6[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{a}}]+2[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{a}}] \\
&-\quad-12[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]+3[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{a}}]-[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{a}}]
\end{aligned}
\]
\[
\begin{aligned}
(\because[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{a}}]=& {[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{a}}]=[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{a}}]=0) } \\
=& 24[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]-[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{a}}] \\
&(\because[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{a}}]) \\
&=24[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]-[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}] \\
&=23[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]
\end{aligned}
\]
Given \(\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}\) are coplanar that is \([\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]=0\)
\[
\text { Hence, } 23 \times 0=0
\]