All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\vec{AB} = \begin{pmatrix}-1\\-8\\1\end{pmatrix} - \begin{pmatrix}7\\8\\9\end{pmatrix} = \begin{pmatrix}--8\\-16\\-8\end{pmatrix}$

$l_{AB}: r = \begin{pmatrix}7\\8\\9\end{pmatrix} + \lambda \begin{pmatrix}1\\2\\1\end{pmatrix}, \lambda \in \mathbb{R}$

(ii)
$\vec{ON} = \begin{pmatrix}{7+\lambda}\\{8+2\lambda}\\{9+\lambda}\end{pmatrix}$ for some $\lambda$.

$\vec{CN} = \begin{pmatrix}{7+\lambda}\\{8+2\lambda}\\{9+\lambda}\end{pmatrix} - \begin{pmatrix}1\\8\\3\end{pmatrix} = \begin{pmatrix}{6+\lambda}\\{2\lambda}\\{6+\lambda}\end{pmatrix}$

Since $\vec{CN}$ is perpendicular to $l_{AB}$

$\Rightarrow \begin{pmatrix}{6+\lambda}\\{2\lambda}\\{6+\lambda}\end{pmatrix} \bullet \begin{pmatrix}1\\2\\1\end{pmatrix} = 0$

$\lambda = -2$

$\therefore, \vec{ON} = \begin{pmatrix}5\\4\\7\end{pmatrix}$

$\vec{AN} = \begin{pmatrix}-2\\-4\\-2\end{pmatrix} = \frac{1}{4} \vec{AB}$

$\therefore, AN : AB = 1: 3$

(iii)
Let C’ be the point of reflection of C in line AB.

$\vec{AC'} = \vec{AC} + 2\vec{CN} = \begin{pmatrix}2\\-8\\2\end{pmatrix}$

$l_{AB}^R: x - 7 = \frac{y-8}{-4} = z-9$