Answer
\[
29=(\vec{w} \times \vec{v})\cdot\vec{u}
\]
Work Step by Step
To evaluate $ (\vec{w} \times \vec{v})\cdot \vec{u}$, we can use the following formula.
\[
(\vec{w} \times \vec{v})\cdot\vec{u}=\left|\begin{array}{lll}
u_{1} & u_{2} & u_{3} \\
v_{1} & v_{2} & v_{3} \\
w_{1} & w_{2} & w_{3}
\end{array}\right|
\]
It's given that
\[
\vec{w}=(-4,1,-3) , \vec{v}=\langle 0,3,2\rangle \text { and } \vec{u}=(1,-2,2)
\]
And then,
\[
\begin{aligned}
(\vec{w} \times \vec{v}) \cdot \vec{u}&=\left|\begin{array}{ccc}
1 & -2 & 2 \\
0 & 3 & 2 \\
-4 & 1 & -3
\end{array}\right| \\
&=0\left|\begin{array}{cc}
-2 & 2 \\
1 & -3
\end{array}\right|+3\left|\begin{array}{cc}
1 & 2 \\
-4 & -3
\end{array}\right|-2\left|\begin{array}{cc}
1 & -2 \\
-4 & 1
\end{array}\right| \\
&=-2(1-8)+3(8-3) \\
&=14+15=29
\end{aligned}
\]
We have found that:
\[
29= (\vec{w} \times \vec{v})\cdot\vec{u}
\]
