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