Answer
\[
\vec{v} \times \vec{u} =\langle-4,-6,-3\rangle
\]
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 $\vec{v}=\langle 3,0,-4\rangle$ and $\vec{u}=\langle 0,1,-2\rangle$, we get that:
\[
\begin{aligned}
\vec{v} \times \vec{u}=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
0 & 1 & -2 \\
3 & 0 & -4
\end{array}\right| &=\hat{i}(-4+0)+\hat{\jmath}(-6+0)+\hat{k}(0-3) \\
&=-4 \hat{\imath}-6 \hat{\jmath}-3 \hat{k} \\
&=\langle-4,-6,-3\rangle
\end{aligned}
\]
Notice that:
\[
(\vec{v} \times \vec{u})\cdot\vec{u} =\langle-4,-6,-3\rangle(0,1,-2) =6-6=0
\]
\[
(\vec{v} \times \vec{u})\cdot\vec{v} =\langle-4,-6,-3\rangle\langle 3,0,-4\rangle=12-12=0
\]
So, $\vec{v}$, $\vec{u}$ are orthogonal to the cross product $\vec{v} \times \vec{u}$
