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

(i)
$cos{\theta}=\frac{|\begin{pmatrix}2\\-2\\1\end{pmatrix}{\bullet}\begin{pmatrix}-6\\3\\2\end{pmatrix}|}{\sqrt{9} \sqrt{49}}=\frac{16}{21}$

$\theta = {40.4}^{\circ}$

(ii) Using Graphing Calculator, equation of line,

$r = \begin{pmatrix}{-\frac{1}{6}}\\{-\frac{2}{3}}\\0\end{pmatrix} + \lambda \begin{pmatrix}7\\10\\6\end{pmatrix}, \lambda \in \mathbb{R}$

(iii)
Distance A to $P_1 = \frac{|\begin{pmatrix}2\\-2\\1\end{pmatrix}{\bullet}\begin{pmatrix}4\\3\\c\end{pmatrix} - 1|}{\sqrt{9}} = \frac{|1+c|}{3}$

Distance A to $P_1 = \frac{|\begin{pmatrix}-6\\3\\2\end{pmatrix}{\bullet}\begin{pmatrix}4\\3\\c\end{pmatrix} + 1|}{\sqrt{49}} = \frac{|2c-14|}{7}$

$\frac{|1+c|}{3} = \frac{|2c-14|}{7}$
$49(1+c)^{2} = 9(2c-14)^{2}$
$c = -49 \mathrm{~or~} \frac{35}{13}$