\((\vec{i}+3\vec{j}-2\vec{k})\times (-\vec{i}+3\vec{k})=\)

The given vectors are \((\vec{i}+3\vec{j}-2\vec{k})\) and \((-\vec{i}+3\vec{k})\).

\((\vec{i}+3\vec{j}-2\vec{k})\times (-\vec{i}+3\vec{k})\) represents the cross product of two vectors.

\((\vec{i}+3\vec{j}-2\vec{k})\times (-\vec{i}+3\vec{k})=\left|\begin{matrix}\vec{i} & \vec{j} & \vec{k} \\ 1 & 3 & -2 \\ -1 & 0 & 3\end{matrix}\right|\\ (\vec{i}+3\vec{j}-2\vec{k})\times (-\vec{i}+3\vec{k})=\vec{i}\left|\begin{matrix}3 & -2 \\ 0 & 3\end{matrix}\right|-\vec{j}\left|\begin{matrix}1 & -2 \\ -1 & 3\end{matrix}\right|+\vec{k}\left|\begin{matrix}1 & 3 \\ -1 & 0\end{matrix}\right|\\ (\vec{i}+3\vec{j}-2\vec{k})\times (-\vec{i}+3\vec{k})=\vec{i}(9-0)-\vec{j}(3-2)+\vec{k}(0+3)\\ (\vec{i}+3\vec{j}-2\vec{k})\times (-\vec{i}+3\vec{k})=9\vec{i}-\vec{j}+3\vec{k}\)

Hence, the value of \((\vec{i}+3\vec{j}-2\vec{k})\times (-\vec{i}+3\vec{k})\) is \(9\vec{i}-\vec{j}+3\vec{k}.\)