Answer
$({\bf u}\times{\bf v}) \cdot{\bf w}= ({\bf v}\times{\bf w}) \cdot {\bf u} = 4$
(the equality stands)
$V=4$
Work Step by Step
${\bf u}\times{\bf v}={\bf i-j+k}\times{\bf 2i+j-2k}=\left|\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k}\\
1 & -1 & 1\\
2 & 1 & -2
\end{array}\right|$
$=(2-1){\bf i}-(-2-2){\bf j}+(1+2){\bf k}= {\bf i}+4{\bf j}+3{\bf k}$
$=\langle 1, 4, 3 \rangle$
${\bf w}={\bf -i+2j-k} = \langle-1, 2, -1 \rangle$
$({\bf u}\times{\bf v}) \cdot {\bf w}=1(-1)+4(2)+3(-1)= 4$
${\bf v}\times{\bf w}={\bf 2i+j-2k}\times{\bf -i+2j-k}=\left|\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k}\\
2 & 1 & -2\\
-1 & 2 & -1
\end{array}\right|$
$=(-1+4){\bf i}-(-2-2){\bf j}+(4+1){\bf k}= 3{\bf i} +4{\bf j}+5{\bf k}$
$=\langle 3, 4, 5 \rangle$
${\bf u}= \langle 1, -1, 1 \rangle$
$({\bf v}\times{\bf w}) \cdot {\bf u}=3(1)+4(-1)+5(1)=4.$
Thus,
$({\bf u}\times{\bf v}) \cdot{\bf w}= ({\bf v}\times{\bf w}) \cdot {\bf u} = 4.$
which is also the volume of the parallelepiped determined by the three vectors,
$V=|({\bf u}\times{\bf v}) \cdot {\bf w}|=4$
