Answer
$a.\quad 2\sqrt{6}$
$b.\displaystyle \quad \frac{2}{\sqrt{6}} {\bf i}+ \frac{1}{\sqrt{6}} {\bf j}+ \frac{1}{\sqrt{6}} {\bf k}$
Work Step by Step
$(a)$
${\bf u}= \overrightarrow{PQ} = \langle 2-1,0+1,-1-2 \rangle= \langle 1,1,-3 \rangle$
${\bf v}= \overrightarrow{PR} = \langle 0-1,2+1,1-2 \rangle= \langle-1,3,-1 \rangle$
We find the area of the parallelogram $|{\bf u}\times{\bf v}|$
The area of the triangle is half the area of the parallelogram.
$A=\displaystyle \frac{1}{2}\cdot|{\bf u}\times{\bf v}|$
${\bf u}\times{\bf v}=\left|\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k}\\
1 & 1 & -3\\
-1 & 3 & -1
\end{array}\right|=(-1+9){\bf i}-(-1-3){\bf j}+(3+1){\bf k}$
$=8{\bf i}+4{\bf j}+4{\bf k}$
$|{\bf u}\times{\bf v}|=\sqrt{64+16+16}=\sqrt{96}=4\sqrt{6}$
$A=\displaystyle \frac{1}{2}\cdot 4\sqrt{6}=2\sqrt{6}$
$(b)$
${\bf w}={\bf u}\times{\bf v}$ is perpendicular to both ${\bf u}$ and ${\bf v}$ (and the plane they belong to).
A unit vector has length 1, so we take
$\displaystyle \frac{{\bf w}}{|{\bf w}|}= \frac{1}{4\sqrt{6}} (8{\bf i}+4{\bf j}+4{\bf k})$
$= \displaystyle \frac{2}{\sqrt{6}} {\bf i}+ \frac{1}{\sqrt{6}} {\bf j}+ \frac{1}{\sqrt{6}} {\bf k}$
