\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 cutting the
$x$ axis at (1,0,0), the $y$ axis at (0,2,0) and the $z$ axis at
(0,0,3).
\medskip

{\bf Answer}

${}$

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\put(3,3){\vector(-1,-1){3}}

\put(3,3){\vector(0,1){4}}

\put(3,6){\line(2,-5){2}}

\put(3,6){\line(-1,-4){1}}

\put(2,2){\line(3,-1){3}}

\put(.25,0){\makebox(0,0)[b]{$x$}}

\put(5.75,0){\makebox(0,0)[b]{$y$}}

\put(3,7){\makebox(0,0)[l]{$z$}}

\put(2,2){\makebox(0,0)[tl]{$A$}}

\put(2,2){\makebox(0,0)[br]{$(1,0,0)$}}

\put(5,1){\makebox(0,0)[tr]{$B$}}

\put(5,1){\makebox(0,0)[bl]{$(0,2,0)$}}

\put(3,6){\makebox(0,0)[r]{$C$}}

\put(3,6){\makebox(0,0)[l]{$(0,0,3)$}}

\put(3,3){\makebox(0,0)[r]{$0$}}

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\put(3.5,4){\line(5,2){1}}

\put(4.45,4.4){\circle*{.125}}

\put(4.6,4.4){\makebox(0,0)[l]{$\hat{\bf{n}}$}}

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\end{picture} \vspace{.5in}

Let \begin{eqnarray*} {\bf{OA}} & = & (1,0,0)={\bf{i}}\\ {\bf{OB}}
& = & (0,2,0)=2{\bf{j}}\\ {\bf{OC}} & = & (0,0,3)=3{\bf{k}}
\end{eqnarray*}

So two vectors in the plane are, e.g.,

${\bf{AB}}$ and ${\bf{BC}}$

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

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

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

Thus a vector equation is

\begin{eqnarray*} {\bf{r}} & = & {\bf{OA}} +
\lambda({\bf{AB}})+\mu({\bf{BC}})\\ & = &
{\bf{i}}+\lambda(-{\bf{i}}+2{\bf{j}})+\mu(-2{\bf{j}}+3{\bf{k}})
\end{eqnarray*}

Cartesian equation is derived as in Q11.

Let unit vector normal to plane $ABC$ be $\hat {\bf{n}}$.

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

\begin{eqnarray*} & & {\bf{AB}}\times{\bf{BC}}\\ & = & \left|\begin{array}{ccc} {\bf{i}} &
{\bf{j}} & {\bf{k}}\\ -1 & 2 & 0\\ 0 & -2 & 3\end{array}\right|\\
& = & {\bf{i}}(2 \times 3)-{\bf{i}}(-2 \times 0)-{\bf{j}}(-1
\times 3)\\ & & +{\bf{k}}(-1 \times -2)+{\bf{j}}(0 \times
0)-{\bf{k}}(0 \times 2)\\ & = & 6{\bf{i}}+3{\bf{j}}+2{\bf{k}}
\end{eqnarray*}

$|{\bf{AB}}\times{\bf{BC}}|=\sqrt{6^2+3^2+2^2}=\sqrt{36+9+4}=\sqrt{49}=7$

so
$\hat{\bf{n}}=\ds\frac{6}{7}{\bf{i}}+\ds\frac{3}{7}{\bf{j}}+\ds\frac{2}{7}{\bf{k}}$

We now need the perpendicular distance, $d$, from 0 to the plane.
This is given by $d={\bf{r}}\cdot\hat{\bf{n}}$ where ${\bf{r}}$ is
the position vector if any point in the plane; say
${\bf{OA}}={\bf{i}}$

Thus $d={\bf{i}} \cdot
\left(\ds\frac{6}{7}{\bf{i}}+\ds\frac{3}{7}{\bf{j}}
+\ds\frac{2}{7}{\bf{k}}\right)=\ds\frac{6}{7}$

Thus we have that $\begin{array} {cl} & (x,y,z) \cdot
\hat{\bf{n}}=d\\ \rm{so} & (x,y,z) \cdot
\left(\ds\frac{6}{7}{\bf{i}}+\ds\frac{3}{7}{\bf{j}}
+\ds\frac{2}{7}{\bf{k}}\right)=\ds\frac{6}{7}\\ \Rightarrow &
\un{6x+3y+2z=6} \end{array}$
\end{document}