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

{\bf Question}

\medskip

{\bf Answer}

\begin{description}
\item[(i)]
$2x+y-2z=3$

${\bf{r}} \cdot \hat{\bf{n}}=d$

$(\undb{2{\bf{i}}+{\bf{j}}-2{\bf{k}}}) \cdot
(\undb{x{\bf{i}}+y{\bf{j}}+z{\bf{k}}})=3$

\hspace{.37in} ${\bf{n}}$ \hspace{.9in} ${\bf{r}}$

$\hat{\bf{n}}=\ds\frac{{\bf{n}}}{|{\bf{n}}|}
=\ds\frac{2{\bf{i}}+{\bf{j}}-2{\bf{k}}}{\sqrt{2^2+1^2+2^2}}
=\ds\frac{2}{3}{\bf{i}}+\ds\frac{1}{3}{\bf{j}}-\ds\frac{2}{3}{\bf{k}}$

Therefore ${\bf{r}} \cdot
\hat{\bf{n}}(x{\bf{i}}+y{\bf{j}}+z{\bf{k}}) \cdot
\left(\ds\frac{2}{3}{\bf{i}}+\ds\frac{1}{3}{\bf{j}}
-\ds\frac{2}{3}{\bf{k}}\right)=\ds\frac{2}{3}=|\hat{\bf{n}}|$

Therefore ${\bf{r}} \cdot
\left(\undb{\ds\frac{2}{3}{\bf{i}}+\ds\frac{1}{3}{\bf{j}}-ds\frac{2}{3}{\bf{k}}}\right)=1$

\hspace{1in} $\hat{\bf{n}}$

$\hat{\bf{n}}$ is the unit normal vector to the plane


${}$

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$d$ is the perpendicular distance from orange $\un{=1}$.

\item[(ii)]
\begin{description}

\item[(a)]
Intersection: Substitute
${\bf{r}}=-4{\bf{i}}+2{\bf{j}}+{\bf{k}}+\lambda(2{\bf{i}}+{\bf{j}}+{\bf{k}})$

into ${\bf{r}} \cdot \hat{\bf{n}}=d$

Therefore
$[(-4+2\lambda){\bf{i}}+(2+\lambda){\bf{j}}+(1+\lambda){\bf{k}}]
\cdot\left[\ds\frac{2}{3}{\bf{i}}+\ds\frac{1}{3}{\bf{j}}-\ds\frac{2}{3}{\bf{k}}\right]=1$

$\Rightarrow
(-4+2\lambda)\ds\frac{2}{3}+(2+\lambda)\ds\frac{1}{3}-(1+\lambda)\ds\frac{2}{3}=1$

$\Rightarrow -8+4\lambda+2+\lambda-2-2\lambda=3$

$\Rightarrow 3 \lambda=11 \Rightarrow
\un{\lambda=\ds\frac{11}{3}}$

Therefore break in line,

$\begin{array}{rcl} {\bf{r}} & = &
-4{\bf{i}}+2{\bf{j}}+{\bf{k}}+\ds\frac{11}{3}(2{\bf{i}}+{\bf{j}}+{\bf{k}})\\
& = &
\ds\frac{10}{3}{\bf{i}}+\ds\frac{17}{3}{\bf{j}}+\ds\frac{14}{3}{\bf{k}}
\end{array}$

\item[(b)]
Substitute
${\bf{r}}=3{\bf{i}}+3{\bf{j}}+2{\bf{k}}+\mu({\bf{i}}+{\bf{k}})$

into ${\bf{r}} \cdot \hat{\bf{n}}=d$

$(3{\bf{i}}+3{\bf{j}}+2{\bf{k}}+\mu{\bf{i}}+\mu{\bf{k}}) \cdot
\left(\ds\frac{2}{3}{\bf{i}}+\ds\frac{1}{3}{\bf{j}}-\ds\frac{2}{3}{\bf{k}}\right)=1$

$(3+\mu)\times\ds\frac{2}{3}+3\times\ds\frac{1}{3}+(2+\mu)\times-\ds\frac{2}{3}=1$

$\Rightarrow 6+2\mu+3-4-2\mu=3$

$\Rightarrow \un{5=3}$!!!

So no value of $\mu$ can be found.

Therefore no intersection.

Therefore line is parallel to plane:

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\end{description}
\end{description}
\end{document}