Find unit vector perpendicular to the plane passing through the points $(1, 2, 3), (2, –1, 1$) and $(1, 2, –4)$

Let ${ }^{\mathrm{\prime }}{O}^{\mathrm{\prime }}$ be the origin and let$A(1, 2, 3), B(2, –1, 1)$ and $C(1, 2, –4)$ be the given points. 

Then $\overline{OA}=\overline{i}+2\overline{j}+3\overline{k},\overline{OB}=2\overline{i}-\overline{j}+\overline{k}$ 

$\overline{OC}=\overline{i}+2\overline{j}-4\overline{k}$

$\overline{AB}=\overline{OB}-\overline{OA}=2\overline{i}-\overline{j}+\overline{k}-\overline{i}-2\overline{j}-3\overline{k}$

$=\overline{i}-3\overline{j}-2\overline{k}$

$\overline{AC}=\overline{OC}-\overline{OA}=\overline{i}+2\overline{j}-4\overline{k}-\overline{i}-2\overline{j}-3\overline{k}=-7\overline{k}$

$\overline{AB}\times \overline{AC}=\left|\begin{array}{ccc}\overline{i}& \overline{j}& \overline{k}\\ 1& -3& -2\\ 0& 0& -7\end{array}\right|=\overline{i}\left(21\right)-\overline{j}\left(-7\right)+\overline{k}\left(0\right)$

$=21\overline{i}+7\overline{j}=7(3\overline{i}+\overline{j})$

$|\overline{AB}\times \overline{AC}|=7\left(\sqrt{3^2+1}\right)=7\sqrt{10}$

Unit vector perpendicular to the plane A, B, C 

$\frac{\overline{AB}\times \overline{AC}}{|\overline{AB}\times \overline{AC}|}=\pm \left(\frac{7(3\overline{i}+\overline{j})}{7\sqrt{10}}\right)=\pm \left(\frac{3\overline{i}+\overline{j}}{\sqrt{10}}\right)$