How would you find a unit vector perpendicular to plane ABC where points are A(3,-1,2),...

We know that, given two vectors, say ${\displaystyle \overrightarrow{x}\&\overrightarrow{y}}$, their Vector
or Outer Product , denoted by ${\displaystyle \overrightarrow{x}\times \overrightarrow{y},}$ is a vector that is
perpendicular to the plane that contains them.
The given pts. ${\displaystyle A(3,-1,2),B(1,-1,-3)\text{and}C(4,-3,1)}$ lie in the plane ${\displaystyle ABC}$
Accordingly, the vectors ${\displaystyle \overrightarrow{AB}\text{and}\overrightarrow{AC}\in \text{the plane}ABC.}$
Thus, ${\displaystyle \overrightarrow{AB}\times \overrightarrow{AC}\perp \text{plane}ABC}$
Therefore, the reqd. unit vector will be
${\displaystyle \frac{\overrightarrow{AB}\times \overrightarrow{AC}}{\left||\overrightarrow{AB}\times \overrightarrow{AC}|\right|}}$
We have, ${\displaystyle \overrightarrow{AB}=(1-3,-1+1,-3-2)=(-2,0,-5)}$,
${\displaystyle \overrightarrow{AC}=(1,-2,-1)}$, so that,
${\displaystyle \overrightarrow{AB}\times \overrightarrow{AC}=\det \left|\begin{array}{ccc}i& j& k\\ -2& 0& -5\\ 1& -2& -1\end{array}\right|}$
${\displaystyle =-10i-7j-4k=(-10,-7,-4)}$
${\displaystyle \Rightarrow \left||\overrightarrow{AB}\times \overrightarrow{AC}|\right|=\sqrt{100+49+16}=\sqrt{165}}$
Finally, the desired unit vector is
${\displaystyle (-\frac{10}{\sqrt{165}},-\frac{7}{\sqrt{165}},-\frac{4}{\sqrt{165}}).}$