Let $u=[3,−2,1]$, $v=[1,1,1]$ and $w=[2,−2,0]$. The parallelepiped formed by $u$ and $v$ and $k(u × v)$ will have the same volume as the parallelepiped formed by $u,v$ and $w$ if $k=±\frac{1}{n}.$
Then what is $n$?
So I have compute the area of base which happens to be $\sqrt{38}$ and the volume is $2$, but how does it connect to the scalar?
The area of the base is $|u\times v|=38$, not $\sqrt{38}$. So the first parallelepiped has volume $|38k(u\times v)|=1444|k|$, because $k(u\times v)$ is perpendicular to the plane containing the base.
And the second paralellepiped has volume $|w.(u\times v)|$, which equals $2$, as you say. And now it's easy.