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

{\bf Question}

Find the line of intersection between the planes $x+2y+3z=6$ and
$x+y+z=1$.

\medskip

{\bf Answer}

${}$

$\left.\begin{array} {crcl} (1) & x+2y+3z & = & 6\\ (2) &  x+y+z &
= & 1 \end{array} \right\}$ require line of intersection: 2 planes
intersect in a line common to both planes

PICTURE \vspace{0.5in}

Take the two simultaneous equations away from each other to
eliminate $x$:

$(1)-(2)$:

$x+2y+3z=6$

$\un{x+\ y+\ z\ =1}$

$\ \ \ \ \ \ \ y+2z=5$

Thus the planes intersect along $y=2z+5$ or \un{$y=-2z+5$}.

But this doesn't tell us about the $x$ dependence.

Thus we have to do another elimination, say of $y$ by $2
\times(2)-(1)$

$2x+2y+2z=\ 2$

$\un{\ x+2y+3z=\ 6}$

$\ x\ \ \ \ \ \ \ -\ z=-4$

Hence we have that

\begin{eqnarray*} y+2z & = & 5\ \ (3)\\ x-z & = & -4\ \ (4)
\end{eqnarray*}

Thus $x+4=z$ from $(4)$

and $-\left(\ds\frac{y+5}{2}\right)=z$ from $(3)$

Remember a straight line in $3-D$ is given by

$$\ds\frac{x-\alpha_1}{\beta_1}=\ds\frac{y-\alpha_2}{\beta_2}
=\ds\frac{z-\alpha_3}{\beta_3}$$

for given $\alpha_i,\ \beta_i$

Thus we combine the above two equalities to get

$$\un{x+4=-\left(\ds\frac{y+5}{2}\right)=z}$$

as the equation of intersection.

Note that from questions above, we can write this in vector form:

$x+4=\lambda;\ -\left(\ds\frac{y+5}{2}\right)=\lambda;\
z=-\lambda$

$\Rightarrow x=\lambda-4;\ y=-5-2\lambda;\ z=\lambda$ parametric
equation

$\Rightarrow
{\bf{r}}=x{\bf{i}}+y{\bf{j}}+z{\bf{k}}=(\lambda-4){\bf{i}}+
(-5-2\lambda){\bf{j}}+\lambda{\bf{k}}$

$\Rightarrow
\un{{\bf{r}}=-4{\bf{i}}-5{\bf{j}}+\lambda({\bf{i}}-2{\bf{j}}+{\bf{k}})}$

\end{document}