Answer
$a.\displaystyle \quad \frac{\sqrt{14}}{2}$
$b.\displaystyle \quad \frac{\sqrt{14}}{7} {\bf i}+ \frac{3\sqrt{14}}{14} {\bf j}+ \frac{\sqrt{14}}{14} {\bf k}$
Work Step by Step
$ \overrightarrow{PQ} = \langle 0+2, 1-2, -1-0 \rangle= \langle 2, -1, -1 \rangle$
$ \overrightarrow{PR} = \langle-1+2, 2-2, -2-0 \rangle= \langle 1, 0, -2 \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}\\
2 & -1 & -1\\
1 & 0 & -2
\end{array}\right|$
$=(2-0){\bf i}-(-4+1){\bf j}+(0+1){\bf k}$
$= 2{\bf i}+3{\bf j}+{\bf k}$
$| \overrightarrow{PQ} \times \overrightarrow{PR}|=\sqrt{4+9+1}=\sqrt{14}$
$A=\displaystyle \frac{1}{2}\cdot\sqrt{14}= \frac{\sqrt{14}}{2}$
$(b)$
${\bf w}= \overrightarrow{PQ} \times \overrightarrow{PR}$ is perpendicular to both $ \overrightarrow{PQ}$ and $ \overrightarrow{PR}$
(and the plane they belong to)
A unit vector has length 1, so we take
$\displaystyle \frac{{\bf w}}{|{\bf w}|}= \frac{1}{\sqrt{14}} (2{\bf i}+3{\bf j}+{\bf k} )$
$= \displaystyle \frac{2}{\sqrt{14}} {\bf i}+ \frac{3}{\sqrt{14}} {\bf j}+ \frac{1}{\sqrt{14}} {\bf k}$
$= \displaystyle \frac{\sqrt{14}}{7} {\bf i}+ \frac{3\sqrt{14}}{14} {\bf j}+ \frac{\sqrt{14}}{14} {\bf k}$
