Answer
\[
45=V
\]
Work Step by Step
If $\vec{v}, \vec{w}$ and $\vec{u}$ are non-zero vectors in 3d-space, then the volume of the shape that has $\vec{v}, \vec{u}$ and $\vec{w}$ as adjacent edges is:
\[
V=| (\vec{w} \times \vec{v})\cdot\vec{u} |=a b s\left|\begin{array}{lll}
u_{1} & u_{2} & u_{3} \\
v_{1} & v_{2} & v_{3} \\
w_{1} & w_{2} & w_{3}
\end{array}\right|
\]
It's given that:
\[
\begin{array}{l}
\vec{w}=\hat{\imath}+2 \hat{\jmath}+4 \hat{k}=\langle 1,2,4\rangle \\
\vec{v}=4 \hat{\imath}+5 \hat{\jmath}+\hat{k}=\langle 4,5,1\rangle \\
\vec{u}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k}=\langle 3,1,2\rangle
\end{array}
\]
And then,
\[
\begin{aligned}
2(-5+8) \vec{u} \cdot(\vec{w} \times \vec{v})+3(-2+20)-(-1+16) &=\left|\begin{array}{ccc}
3 & 1 & 2 \\
4 & 5 & 1 \\
1 & 2 & 4
\end{array}\right| \\
&=3\left|\begin{array}{cc}
5 & 1 \\
2 & 4
\end{array}\right|-\left|\begin{array}{cc}
4 & 1 \\
1 & 4
\end{array}\right|+2\left|\begin{array}{cc}
4 & 5 \\
1 & 2
\end{array}\right| \\
&=-(16-1)+(-5+8)2+(-2+20) 3\\
&45=-15+6+54
\end{aligned}
\]
So, the volume of the parallepiped is:
\[
45 =V
\]
