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