\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Find the Cartesian and vector equations of the plane passing
through the points (1,1,1),(1,2,3) and (3,2,1).

\medskip

{\bf Answer}

Let \begin{eqnarray*} {\bf{OA}} & = &
(1,1,1)={\bf{i}}+{\bf{j}}+{\bf{k}}\\ {\bf{OB}} & = &
(1,2,3)={\bf{i}}+2{\bf{j}}+3{\bf{k}}\\ {\bf{OC}} & = &
(3,2,1)=3{\bf{i}}+2{\bf{j}}+{\bf{k}} \end{eqnarray*}

Need two vectors in the plane, say ${\bf{AB}},\ {\bf{BC}}$

${\bf{AB}}={\bf{AO}}+{\bf{OB}}=-{\bf{i}}-{\bf{j}}-
{\bf{k}}+{\bf{i}}+2{\bf{j}}+3{\bf{k}}={\bf{j}}+2{\bf{k}}$

${\bf{BC}}={\bf{BO}}+{\bf{OC}}=-{\bf{i}}-2{\bf{j}}-
3{\bf{k}}+3{\bf{i}}+2{\bf{j}}+{\bf{k}}=2{\bf{i}}-2{\bf{k}}$

A  vector to a point in the plane is
${\bf{OA}}={\bf{i}}+{\bf{j}}+{\bf{k}}$

Thus a vector equation is

\begin{eqnarray*} {\bf{r}} & = & {\bf{OA}} +
\lambda({\bf{AB}})+\mu({\bf{BC}})\\ \Rightarrow & = &
\un{{\bf{i}}+{\bf{j}}+{\bf{k}}+\lambda({\bf{j}}+2{\bf{k}})+\mu(2{\bf{i}}-2{\bf{k}})}
\end{eqnarray*}

We now need $\hat{\bf{n}}$ a vector normal to plane $ABC$

$$\hat{\bf{n}}=\ds\frac{{\bf{AB}} \times {\bf{BC}}}{|{\bf{AB}}
\times {\bf{BC}}|}$$

\begin{eqnarray*} & & {\bf{AB}}\times{\bf{BC}}\\ & = & \left|\begin{array}{ccc} {\bf{i}} &
{\bf{j}} & {\bf{k}}\\ 0 & 1 & 2\\ 2 & 0 & -2\end{array}\right|\\ &
= & {\bf{i}}(1 \times -2)-{\bf{i}}(0 \times 2)-{\bf{j}}(0 \times
-2)\\ & & +{\bf{k}}(0 \times 0)+{\bf{j}}(2 \times 2)-{\bf{k}}(2
\times 1)\\ & = & -2{\bf{i}}+4{\bf{j}}-2{\bf{k}}
\end{eqnarray*}

$|{\bf{AB}}\times{\bf{BC}}|=\sqrt{(-2)^2+4^2+(-2)^2}=\sqrt{4+16+4}=\sqrt{24}$

so
$\hat{\bf{n}}=-\ds\frac{2}{\sqrt{24}}{\bf{i}}+\ds\frac{4}{\sqrt{24}}{\bf{j}}
-\ds\frac{2}{\sqrt{24}}{\bf{k}}=-\ds\frac{{\bf{i}}}{\sqrt{6}}+2\ds\frac
{{\bf{j}}}{\sqrt{6}}-\ds\frac{{\bf{k}}}{\sqrt{6}}$

The perpendicular distance $d$ from 0 to the plane is $d={\bf{r}}
\cdot \hat{\bf{n}}$ where ${\bf{r}}$ is the position vector of any
point in the plane, say ${\bf{OA}}={\bf{i}}+{\bf{j}}+{\bf{k}}$.
Thus

$$d=({\bf{i}}+{\bf{j}}+{\bf{k}}) \cdot
\left(-\ds\frac{{\bf{i}}}{\sqrt{6}}+2\ds\frac{{\bf{j}}}
{\sqrt{6}}-\ds\frac{{\bf{k}}}{\sqrt{6}}\right)=
-\ds\frac{1}{\sqrt{6}}+2\ds\frac{2}{\sqrt{6}}-\ds\frac{1}{\sqrt{6}}=\un{0}$$

Thus the plane actually passes through the origin!

Thus we have for a general point $(x,y,z)={\bf{r}}$,

$(x,y,z) \cdot \left(-\ds\frac{1}{\sqrt{6}},\
\ds\frac{2}{\sqrt{6}},\ -\ds\frac{1}{\sqrt{6}}\right)=0$

$\Rightarrow -x+2y-z=0$

$\rm{or} \un{x-2y+x=0}$

\end{document}