Answer
\[
\vec{v} \times \vec{u}=\langle 7,10,9\rangle
\]
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 $\vec{v}=\langle-4,1,2\rangle$ and $\vec{u}=\langle 1,2,-3\rangle$, we get that:
\[
\begin{aligned}
\vec{v} \times \vec{u} &=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
1 & 2 & -3 \\
-4 & 1 & 2
\end{array}\right|=\hat{\imath}(4+3)-\hat{\jmath}(2-12)+\hat{k}(1+8) \\
&=7 \hat{\imath}+10 \hat{\jmath}+9 \hat{k} \\
&=\langle 7,10,9\rangle
\end{aligned}
\]
Notice that
\[
\begin{array}{l}
(\vec{v} \times \vec{u})\cdot\vec{u} =7,10,9\rangle(1,2,-3)\langle =-27+7+20=0 \\
(\vec{v} \times \vec{u})\cdot\vec{v} =\langle 7,10,9\rangle\langle-4,1,2\rangle=10+18-28=0
\end{array}
\]
