Answer
$a.\quad \sqrt{14}$
$b.\quad 1$
Work Step by Step
$ a.\quad$
The area of the paralellogram equals $|{\bf u}\times{\bf v}|$
${\bf u}\times{\bf v}=\left|\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k}\\
1 & 1 & -1\\
2 & 1 & 1
\end{array}\right|$
$=(1+1){\bf i}-(1+2){\bf j}+(1-2){\bf k}$
$=2{\bf i}-3{\bf j-k}$
Area = $|{\bf u}\times{\bf v}| = \sqrt{4+9+1}=\sqrt{14}$
$ b.\quad$
Volume = $({\bf u}\times{\bf v} )\cdot{\bf w}\qquad $(triple scalar product)
We can calculate this as a determinant, but we already have ${\bf u}\times{\bf v}$, so
Volume $=2(-1)-3(-2)+(-1)(3)$
$=1$
