Answer
\[
16 =V
\]
Work Step by Step
If $\vec{w}, \vec{u}$ and $\vec{v}$ are nonzero vectors in 3d-space, then the volume of the shape with $\vec{u}, \vec{v}$ and $\vec{w}$ as contiguous edges is:
\[
V=|(\vec{w} \times \vec{v}) \cdot \vec{u}|=a b s\left|\begin{array}{ccc}
u_{1} & u_{2} & u_{3} \\
v_{1} & v_{2} & v_{3} \\
w_{1} & w_{2} & w_{3}
\end{array}\right|
\]
It's given that:
\[
\vec{w}=\langle 2,2,-4\rangle, \vec{v}=\langle 0,4,-2\rangle \text { and } \vec{u}=\langle 2,-6,2\rangle
\]
And then
\[
\begin{aligned}
(\vec{w} \times \vec{v}) \cdot\vec{u} &=\left|\begin{array}{ccc}
2 & -6 & 2 \\
0 & 4 & -2 \\
2 & 2 & -4
\end{array}\right| \\
&=-0\left|\begin{array}{cc}
-6 & 2 \\
2 & -4
\end{array}\right|+4\left|\begin{array}{cc}
2 & 2 \\
2 & -4
\end{array}\right|+2\left|\begin{array}{cc}
2 & -6 \\
2 & 2
\end{array}\right| \\
&=(12+4)2+(-4-8)4 \\
&-16=32-48
\end{aligned}
\]
So, the volume is:
\[
16 =|-16|=V
\]
