Answer
\[\left\langle -1,-1,-1 \right\rangle \]
Work Step by Step
\[\begin{align}
& \text{Let the vectors: }\mathbf{u}=\left\langle 2,-3,1 \right\rangle ,\text{ }\mathbf{v}=\left\langle 1,-2,1 \right\rangle \\
& \\
& \text{Find }\mathbf{u}\times \mathbf{v} \\
& \mathbf{u}\times \mathbf{v}=\left| \begin{matrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & -3 & 1 \\
1 & -2 & 1 \\
\end{matrix} \right| \\
& \mathbf{u}\times \mathbf{v}=\left| \begin{matrix}
-3 & 1 \\
-2 & 1 \\
\end{matrix} \right|\mathbf{i}-\left| \begin{matrix}
2 & 1 \\
1 & 1 \\
\end{matrix} \right|\mathbf{j}+\left| \begin{matrix}
2 & -3 \\
1 & -2 \\
\end{matrix} \right|\mathbf{k} \\
& \mathbf{u}\times \mathbf{v}=-\mathbf{i}-\mathbf{j}-\mathbf{k} \\
& \mathbf{u}\times \mathbf{v}=\left\langle -1,-1,-1 \right\rangle \\
& \\
& \text{Now, show that }\mathbf{u}\times \mathbf{v}\text{ is orthogonal to }\mathbf{u}\text{ and }\mathbf{v} \\
& \mathbf{u}\cdot \left( \mathbf{u}\times \mathbf{v} \right)=\left\langle 2,-3,1 \right\rangle \cdot \left\langle -1,-1,-1 \right\rangle \\
& \mathbf{u}\cdot \left( \mathbf{u}\times \mathbf{v} \right)=-2+3-1 \\
& \mathbf{u}\cdot \left( \mathbf{u}\times \mathbf{v} \right)=0\to \mathbf{u}\bot \left( \mathbf{u}\times \mathbf{v} \right) \\
& \\
& \mathbf{v}\cdot \left( \mathbf{u}\times \mathbf{v} \right)=\left\langle 1,-2,1 \right\rangle \cdot \left\langle -1,-1,-1 \right\rangle \\
& \mathbf{v}\cdot \left( \mathbf{u}\times \mathbf{v} \right)=-1+2-1 \\
& \mathbf{v}\cdot \left( \mathbf{u}\times \mathbf{v} \right)=0\to \mathbf{u}\bot \left( \mathbf{u}\times \mathbf{v} \right) \\
& \\
& \mathbf{u}\times \mathbf{v}=\left\langle -1,-1,-1 \right\rangle \\
\end{align}\]
