Answer
\[
80=(\vec{w} \times \vec{v})\cdot\vec{u}
\]
Work Step by Step
To evaluate $ (\vec{w} \times \vec{v})\cdot \vec{u}$, we can use the following formula.
\[
(\vec{w} \times \vec{v})\cdot\vec{u} =\left|\begin{array}{ccc}
u_{1} & u_{2} & u_{3} \\
v_{1} & v_{2} & v_{3} \\
w_{1} & w_{2} & w_{3}
\end{array}\right|
\]
It's given that
\[
\begin{array}{l}
\vec{w}=\hat{\jmath}+5 \hat{k}=\langle 0,1,5\rangle\\
\vec{v}=4 \hat{\imath}+\hat{\jmath}-3 \hat{k}=\langle 4,1,-3\rangle \\
\vec{u}=2 \hat{\imath}-3 \hat{\jmath}+\hat{k}=\langle 2,-3,1\rangle
\end{array}
\]
And then
\[
\begin{aligned}
(\vec{w} \times \vec{v})\cdot \vec{u} &=\left|\begin{array}{ccc}
2 & -3 & 1 \\
4 & 1 & -3 \\
0 & 1 & 5
\end{array}\right| \\
&=0\left|\begin{array}{cc}
-3 & 1 \\
1 & -3
\end{array}\right|-1\left|\begin{array}{cc}
2 & 1 \\
4 & -3
\end{array}\right|+5\left|\begin{array}{cc}
2 & -3 \\
4 & 1
\end{array}\right| \\
&=5(2+12)-(-6-4) \\
&=70+10=80
\end{aligned}
\]
We get that
\[
80=(\vec{w} \times \vec{v})\cdot\vec{u}
\]
