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

{\bf Question}

Find the intersection of the line $x=1+3t,\ y=2+4t,\ z=3+5t$ with
the plane $x+2y+3z=6$.

\medskip

{\bf Answer}

${}$

$\left.\begin{array}{l} x=1+3t\\y=2+4t\\z=3+5t
\end{array}\right\}$ intersection with $\left\{\begin{array}
{l}x+2y+3z=6\\ {}\\ {} \end{array} \right.$

The intersection occurs when

$\begin{array} {cl} &(1+3t)+2(2+4t)+3(3+5t)=6\\ \Rightarrow &
1+3t+4+8t+9+15t=6\\ \Rightarrow 14+26t=6\\ \Rightarrow
t=\ds\frac{6-14}{26}=-\ds\frac{8}{26}=-\ds\frac{4}{13}
\end{array}$

Thus the line intersects the plane when $t=-\ds\frac{4}{13}$

$\Rightarrow \begin{array} {l}
x=1+3\left(-\ds\frac{4}{13}\right)=\ds\frac{1}{13}\\
y=2+4\left(-\ds\frac{4}{13}\right)=\ds\frac{10}{13}\\
z=3+5\left(-\ds\frac{4}{13}\right)=\un{\ds\frac{19}{13}}
\end{array}$

\end{document}
