Answer
\[
\vec{v} \times \vec{u}=\langle-1,-2,-7\rangle
\]
Work Step by Step
For two 3 -Dimensional vectors
\[
\vec{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle \quad \text { and } \quad\vec{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle
\]
The cross product is given by the determinant
\[
\vec{v} \times \vec{u}=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
u_{1} & u_{2} & u_{3} \\
v_{1} & v_{2} & v_{3}
\end{array}\right|
\]
And then, for $\vec{u}=3 \hat{\imath}+2 \hat{\jmath}-\hat{k}=\langle 3,2,-1\rangle$ and $\vec{v}=-\hat{\imath}-3 \hat{\jmath}+\hat{k}=\langle-1,-3,1\rangle$, we get:
\[
\begin{aligned}
\vec{v} \times \vec{u} &=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
3 & 2 & -1 \\
-1 & -3 & 1
\end{array}\right|=\hat{\imath}(2-3)+\hat{\jmath}(1-3)+\hat{k}(-9+2) \\
&=-\hat{\imath}-2 \hat{\jmath}-7 \hat{k}=\langle-1,-2,-7\rangle
\end{aligned}
\]
Notice that
\[
\begin{array}{c}
(\vec{v} \times \vec{u})\cdot\vec{u} =\langle-1,-2,-7\rangle\langle 3,2,-1\rangle=-4+7-3=0 \\
(\vec{v} \times \vec{u})\cdot\vec{v} =\langle-1,-2,-7\rangle \langle-1,-3,1\rangle \\
=6-7+1=0
\end{array}
\]
