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

{\bf Question}

Do the planes $x+2y+3z=6$ and $3x+6y+9z=12$ intersect? How about
$x+2y+3z=6$ and $3x+6y+9z=18$?

\medskip

{\bf Answer}

${}$

$\begin{array}{rcl} x+2y+3z & = & 6\ \ (1)\\ 3x+6y+9z & = & 12\ \
(2)
\end{array}$

Consider $3 \times (1)$: $3x+6y+9z+18\ (3)$

Clearly the LHS of $(3)$=LHS of $(2)$.

They can only now intersect if the RHS of $(3)$ and $(2)$ are also
equal, which they're not. Thus the planes don't intersect.
Actually they're parallel, as can be deduced from the vector form
of the planes:

\setlength{\unitlength}{.25in}
\begin{picture}(4,6)

\put(1,0){\line(1,0){3}}

\put(1,0){\line(-1,2){1}}

\put(4,0){\line(-1,2){1}}

\put(0,2){\line(1,0){3}}

\put(2,1){\circle*{.125}}

\put(2,1){\vector(0,1){2.5}}

\put(2,1){\line(1,0){.4}}

\put(1,3){\line(1,0){3}}

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

\put(4,3){\line(-1,2){1}}

\put(0,5){\line(1,0){3}}

\put(1,4.5){\circle*{.125}}

\put(1,4.5){\vector(0,1){1.5}}

\put(1,4.5){\line(1,0){.4}}

\put(2.2,1){\line(0,1){.2}}

\put(2,1.2){\line(1,0){.2}}

\put(1.2,4.5){\line(0,1){.2}}

\put(1,4.7){\line(1,0){.2}}

\put(1,6){\makebox(0,0)[l]{$\ \hat{\bf{n}}$}}

\put(2,3.5){\makebox(0,0)[l]{$\ \hat{\bf{n}}$}}
\end{picture}
$\begin{array} {l} {\bf{r}} \cdot
(3{\bf{i}}+6{\bf{j}}+9{\bf{k}})=18\\ {\bf{r}} \cdot
(3{\bf{n}}+6{\bf{n}}+9{\bf{n}})=12\\
{\bf{r}}=x{\bf{i}}+y{\bf{j}}+z{\bf{k}} \end{array}$

Clearly they have a common normal vector ${\bf{n}}$. Hence they're
parallel.

The two planes

$\begin{array} {rcl} x+2y+3z & = & 6\ \ (4)\\ 3x+6y+9z & = & 18\ \
(5)
\end{array}$

are coincident/identical as can be seen by multiplying $(4)$
through by 3 to get $(5)$.

\end{document}
