If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are mutually perpendicular vectors having magnitude $1,2,3$ respectively, then $\left[\begin{array}{ccc}\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}& \overrightarrow{b}-\overrightarrow{a}& \overrightarrow{c}\end{array}\right]=$

Given, \(\left.\right|_a \rightarrow|=1,|\vec{b}|=2,|\vec{c}|=3\) and
\(\vec{a} \cdot \vec{b}=0=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}\) (as the three vectors are mutually perpendicular)
So,
\([(\vec{a}+\vec{b}+\vec{c}) \times(\vec{b}-\vec{a})] \cdot \vec{c}\)
\(=[\vec{a} \times \vec{b}-0+0-\vec{b} \times \vec{a}+\vec{c} \times \vec{b}-\vec{c} \times a \vec{a}] \cdot \vec{c}\)
\(=[2(\vec{a} \times \vec{b}) \cdot \vec{c}+(\vec{c} \times \vec{b}) \cdot \vec{c}-(\vec{c} \times \vec{a}) \cdot \vec{c}]\)
\(=2(\vec{a} \times \vec{b}) \cdot \vec{c}+0-0\)
\(=2\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]\)
\(=2 \cdot 1 \cdot 2 \cdot 3\)
\(=12\)