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