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
$(a)$
${\bf u}= \overrightarrow{PQ} = \langle 0+2, 1-2, -1-0 \rangle= \langle 2, -1, -1 \rangle$
${\bf v}= \overrightarrow{PR} = \langle-1+2, 2-2, -2-0 \rangle= \langle 1, 0, -2 \rangle$
We find the area of the parallelogram $|{\bf u}\times{\bf v}|$
The area of the triangle is half the area of the parallelogram.
$A=\displaystyle \frac{1}{2}\cdot|{\bf u}\times{\bf v}|$
${\bf u}\times{\bf v}=\left|\begin{array}{rrr}
{\bf i} & {\bf j} & {\bf k}\\
2 & -1 & -1\\
1 & 0 & -2
\end{array}\right|=(2-0){\bf i}-(-4+1){\bf j}+(0+1){\bf k}$
$= 2{\bf i}+3{\bf j}+{\bf k}$
$|{\bf u}\times{\bf v}|=\sqrt{4+9+1}=\sqrt{14}$
$A=\displaystyle \frac{1}{2}\cdot\sqrt{14}= \frac{\sqrt{14}}{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
$\displaystyle \frac{{\bf w}}{|{\bf w}|}= \frac{1}{\sqrt{14}} (2{\bf i}+3{\bf j}+{\bf k} )$
$= \displaystyle \frac{2}{\sqrt{14}} {\bf i}+ \frac{3}{\sqrt{14}} {\bf j}+ \frac{1}{\sqrt{14}} {\bf k}$
$= \displaystyle \frac{\sqrt{14}}{7} {\bf i}+ \frac{3\sqrt{14}}{14} {\bf j}+ \frac{\sqrt{14}}{14} {\bf k}$
