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