Answer
$a.\quad 3$
$b.\displaystyle \quad \frac{2}{3} {\bf i}+ \frac{2}{3} {\bf j}- \frac{1}{3} {\bf k}$
Work Step by Step
${\bf u}= \overrightarrow{PQ} = \langle 2-1,1-1,3-1 \rangle= \langle 1,0,2 \rangle$
${\bf v}= \overrightarrow{PR} = \langle 3-1,-1-1,1-1 \rangle= \langle 2,-2,0 \rangle$
We find the area of the parallelogram $|{\bf u}\times{\bf v}|$
The area of the triangle is half the area of the parallelogram.
$A=\displaystyle \frac{1}{2}\cdot|{\bf u}\times{\bf v}|$
${\bf u}\times{\bf v}=\left|\begin{array}{lll}
{\bf i} & {\bf j} & {\bf k}\\
1 & 0 & 2\\
2 & -2 & 0
\end{array}\right|=(0+4){\bf i}-(0-4{\bf j}+(-2-0){\bf k}$
$=4{\bf i}+4{\bf j}-2{\bf k}$
$|{\bf u}\times{\bf v}|=\sqrt{16+16+4}=\sqrt{36}=6$
$A=\displaystyle \frac{1}{2}\cdot 6=3$
$(b)$
${\bf w}={\bf u}\times{\bf v}$ is perpendicular to both ${\bf u}$ and ${\bf v}$ (and the plane they belong to).
A unit vector has length 1, so we take
$\displaystyle \frac{{\bf w}}{|{\bf w}|}= \frac{1}{6} (4{\bf i}+4{\bf j}-2{\bf k})$
$= \displaystyle \frac{2}{3} {\bf i}+ \frac{2}{3} {\bf j}- \frac{1}{3} {\bf k}$
