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

(i)
$\theta = \mathrm{cos^{-1}} |\frac{\begin{pmatrix}2\\3\\{-6}\end{pmatrix}\cdot \begin{pmatrix}1\\0\\{0}\end{pmatrix}}{\sqrt{49} \cdot 1}| = 73.4^{\circ}$

(ii)
Let $\vec{ON}$ be point on L that makes $\sqrt{33}$ from P.

$\vec{ON} = \begin{pmatrix}1\\{-2}\\{-4}\end{pmatrix} + \lambda \begin{pmatrix}2\\3\\{-6}\end{pmatrix}$ for some $\lambda$

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

$|\vec{PN}| = \sqrt{(-1+2\lambda)^2 + (-7+3\lambda)^2 + (2-6\lambda)^2}$

$33 = 49 \lambda^2 - 70 \lambda + 54$

$\lambda = 1 \mathrm{~or~} \frac{3}{7}$

$\vec{ON} = \begin{pmatrix}3\\{1}\\{-10}\end{pmatrix} \mathrm{~or~} \frac{1}{7}\begin{pmatrix}13\\{-5}\\{-46}\end{pmatrix}$

$L = |\vec{PN}|^2 = (-1+2\lambda)^2 + (-7+3\lambda)^2 + (2-6\lambda)^2 = 49 \lambda^2 - 70 \lambda +54$

$\frac{dL}{d\lambda} = 98 \lambda - 70$

$\frac{d^2L}{d\lambda ^2} = 98 > 0$

So when $\lambda = \frac{70}{98} = \frac{5}{7}$, L is minimum.

$\vec{ON} = \frac{1}{7} \begin{pmatrix}{17}\\{1}\\{-58}\end{pmatrix}$

(iii)
$\begin{pmatrix}{-1}\\{-7}\\{2}\end{pmatrix} \times \begin{pmatrix}{2}\\{3}\\{-6}\end{pmatrix} = \begin{pmatrix}{36}\\{-2}\\{11}\end{pmatrix}$

$\begin{pmatrix}{1}\\{-2}\\{-4}\end{pmatrix} \bullet \begin{pmatrix}{36}\\{-2}\\{11}\end{pmatrix} = -4$

$\therefore, \pi : 36x - 2y +11z=-4$