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

(i)

Direction vector of $l \text{~is~} \begin{pmatrix}{-3}\\6\\9\end{pmatrix} = -3 \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}$

Normal vector of $p \text{~is~} \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}$

Since $l$ is parallel to the normal vector of $p, l$ is perpendicular to $p$.

(ii)
$l: r = \begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}$

$[\begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix}] \bullet \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix} = 0$

$10 + \lambda + 2 4 \lambda + 9 + 9 \lambda = 0$

$\lambda = - \frac{3}{2}$

Point of intersection $= (\frac{17}{2}, 2, \frac{3}{2})$.

(iii)
$\begin{pmatrix}10\\{-1}\\{-3}\end{pmatrix} + \lambda \begin{pmatrix}1\\{-2}\\{-3}\end{pmatrix} = \begin{pmatrix}{-2}\\23\\33\end{pmatrix}$

$10 + \lambda = - 2$ — (1)

$-1 - 2 \lambda = 23$ — (2)

$-3 - 3 \lambda = 33$ — (3)

Since $\lambda = -12$ satisfies all 3 equations, A lies on l.

Since $= (\frac{17}{2}, 2, \frac{3}{2})$ is the midpoint of A & B, by ratio theorem,

$\begin{pmatrix}{\frac{17}{2}}\\2\\{\frac{3}{2}}\end{pmatrix} = \frac{\vec{OA} + \vec{OB}}{2}$

$\vec{OB} = 2 \begin{pmatrix}{\frac{17}{2}}\\2\\{\frac{3}{2}}\end{pmatrix} - \begin{pmatrix}{-2}\\23\\33\end{pmatrix} = \begin{pmatrix}19\\{-19}\\{-30}\end{pmatrix}$

$B(19, -19, -30)$

(iv)
Area

$= \frac{1}{2} |\vec{OA} \times \vec{OB}|$

$= \frac{1}{2} |\begin{pmatrix}{-2}\\23\\33\end{pmatrix} \times \begin{pmatrix}19\\{-19}\\{-30}\end{pmatrix}|$

$= \frac{1}{2} |\begin{pmatrix}{-63}\\{567}\\{-399}\end{pmatrix}|$

$= \frac{1}{2} \sqrt{63^2 +567^2 + 399^2}$

$= 348$ to the nearest whole number.