\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)]

${}$

\setlength{\unitlength}{.5in}
\begin{picture}(7,4)

\put(0,1){\vector(1,0){5}}

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

\put(0,1){\vector(3,1){6}}

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

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

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

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

\put(5,1.1){\makebox(0,0)[l]{$C$}}

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

\put(2,.5){\makebox(0,0)[l]{$5{\bf{i}}-{\bf{j}}-6{\bf{k}}$}}

\put(3,3){\makebox(0,0)[r]{$2{\bf{i}}-{\bf{j}}+3{\bf{k}}\ $}}

\put(3,2){\makebox(0,0)[l]{${\bf{i}}-{\bf{j}}+{\bf{k}}$}}

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

\put(5,1.5){\makebox(0,0){$\psi$}}

\put(3.5,2.8){\makebox(0,0){$\theta$}}

\put(5.5,2.5){\makebox(0,0){$\phi$}}
\end{picture}

\item[(ii)]
${\bf{AB}}=-{\bf{OA}}+{\bf{OB}}=-2{\bf{i}}+{\bf{j}}-3{\bf{k}}
+{\bf{i}}-{\bf{j}}+{\bf{k}}=-{\bf{i}}-2{\bf{k}}$

${\bf{AC}}=-{\bf{OA}}+{\bf{OC}}=-2{\bf{i}}+{\bf{j}}-3{\bf{k}}
+5{\bf{i}}-{\bf{j}}-6{\bf{k}}=+3{\bf{i}}-9{\bf{k}}$

${\bf{AB}} \cdot {\bf{AC}}=|{\bf{AB}}||{\bf{AC}}|\cos\theta$

so
$\cos\theta=\ds\frac{{\bf{AB}}\cdot{\bf{AC}}}{|{\bf{AB}}||{\bf{AC}}|}$

$|{\bf{AB}}|=\sqrt{1^2+0^2+2^2}=\sqrt{5}$

$|{\bf{AC}}|=\sqrt{3^2+0^2+9^2}=\sqrt{90}$

Therefore

\begin{eqnarray*} \cos\theta & = & \ds\frac{(-{\bf{i}}-2{\bf{k}})\cdot
(3{\bf{i}}-9{\bf{k}})}{\sqrt{5}\sqrt{90}}\\ & = &
\ds\frac{(-1\times 3+0+2\times 9)}{\sqrt{5}\sqrt{90}}\\ & = &
\ds\frac{15}{\sqrt{5}\sqrt{90}}\\ & = & 0.707 \end{eqnarray*}

$\Rightarrow
\theta=cos^{-1}\left(\ds\frac{15}{\sqrt{5}\sqrt{90}}\right) =
45^{\circ}$


${\bf{BA}}=-{\bf{AB}}={\bf{i}}+2{\bf{k}}$

${\bf{BC}}=-{\bf{OB}}+{\bf{OC}}=-{\bf{i}}+{\bf{j}}-{\bf{k}}
+5{\bf{i}}-{\bf{j}}-6{\bf{k}}=4{\bf{i}}-7{\bf{k}}$

${\bf{BA}} \cdot {\bf{BC}}=|{\bf{BA}}||{\bf{BC}}|\cos\phi$

so
$\cos\phi=\ds\frac{{\bf{BA}}\cdot{\bf{BC}}}{|{\bf{BA}}||{\bf{BC}}|}$

$|{\bf{BA}}|=|{\bf{AB}}|=\sqrt{5}$

$|{\bf{BC}}|=\sqrt{4^2+0^2+7^2}=\sqrt{65}$

Therefore

\begin{eqnarray*} \cos\phi & = & \ds\frac{({\bf{i}}+2{\bf{k}})\cdot
(4{\bf{i}}-7{\bf{k}})}{\sqrt{5}\sqrt{65}}\\ & = &
\ds\frac{(4\times 1+0+7\times 2)}{\sqrt{5}\sqrt{65}}\\ & = &
\ds\frac{-10}{\sqrt{5}\sqrt{65}}\\ & = & -0.5547 \end{eqnarray*}

$\Rightarrow \phi=cos^{-1}(-0.5547) = 123.69^{\circ}$

Therefore $\psi=180-\theta-\phi=180-45-123.69=11.3^{\circ}$

\item[(iii)]
\begin{eqnarray*}\rm{Area\ of\ triangle}\ ABC & = & \ds\frac{1}{2}||{\bf{AB}}| \times
|{\bf{AC}}||\ \rm{(e.g)}\\ & = &
\ds\frac{1}{2}\left|\left|\begin{array}{ccc} {\bf{i}} & {\bf{j}} &
{\bf{k}}\\ -1 & 0 & -2\\ 3 & 0 & -9
\end{array}\right|\right|\\ & = &
\ds\frac{1}{2}\left|{\bf{i}}\left|\begin{array}{cc} 0 & -2\\ 0 &
-9 \end{array}\right|-{\bf{j}}\left|\begin{array}{cc} -1 & -2\\ 3
& -9
\end{array}\right|\right.\\ & & \left.+{\bf{k}}\left|\begin{array}{cc} -1 & 0\\ 3 &
0 \end{array}\right|\right|\\ & = &
\ds\frac{1}{2}\left|\left|\begin{array}{cc} -1 & -2\\ +3 & -9
\end{array}\right|\right|\\ & = &
\left|\ds\frac{1}{2}(9+6)\right|\\ & = & \un{\ds\frac{15}{2}}
\end{eqnarray*}

\end{description}
\end{document}
