The point A is (0,7) and the point B is (6,9).

1. Subtract the coordinates of A from the coordinates of B to get the column vector for $\overrightarrow{AB}$.

$A=(0,7)$ and $B=(6,9)$.

$\overrightarrow{AB}=\begin{pmatrix}6-0\\9-7\end{pmatrix}=\begin{pmatrix}6\\2\end{pmatrix}$

1. Find the gradient of line $AB$ using the coordinates of $A$ and $B$.

For $A(0,7)$ and $B(6,9)$,
$m=\frac{9-7}{6-0}=\frac{2}{6}=\frac{1}{3}.$

2. Form the equation of $AB$ and rewrite it in the given form to identify $p$ and $q$.

Using $A(0,7)$ and $m=\frac13$,
$y-7=\frac13(x-0)$
$3y-21=x$
$x-3y+21=0.$
Comparing with $x+py+q=0$,
$p=-3,\quad q=21.$

1. Determine how far beyond C the point M is, using the ratio condition.

Since M is on $BC$ produced, the points are in the order $B, C, M$.

Given $BM=3CM$ and $BM=BC+CM$,
$BC+CM=3CM$
$BC=2CM$
$CM=\frac12 BC$

2. Use the vector from B to C to extend beyond C and find the coordinates of M.

$\overrightarrow{BC}=\begin{pmatrix}12-6\\2-9\end{pmatrix}=\begin{pmatrix}6\\-7\end{pmatrix}$

So
$\overrightarrow{CM}=\frac12\overrightarrow{BC}=\begin{pmatrix}3\\-\frac72\end{pmatrix}$

Hence
$M=(12,2)+\left(3,-\frac72\right)$
$M=\left(15,-\frac32\right)$

1. Apply the distance formula using the coordinates of A and C.

A is $(0,7)$ and C is $(12,2)$.

$AC=\sqrt{(12-0)^2+(2-7)^2}$
$=\sqrt{144+25}$
$=\sqrt{169}=13$

1. Use the fact that $CD$ is parallel to the $y$-axis to identify the $x$-coordinate of $D$.

Since $CD$ is parallel to the $y$-axis, $C$ and $D$ have the same $x$-coordinate.

Given $C=(12,2)$,
$x_D=12.$

2. Substitute $x=12$ into the equation of line $AB$ to find the $y$-coordinate of $D$.

From part (b), the equation of $AB$ is
$x-3y+21=0.$

Substitute $x=12$:
$12-3y+21=0$
$33-3y=0$
$y=11.$

Therefore,
$D=(12,11).$

1. Find \(\overrightarrow{AD}\) and compare it with \(\overrightarrow{AB}\).

From earlier parts,
$A=(0,7),\quad D=(12,11),\quad \overrightarrow{AB}=\begin{pmatrix}6\\2\end{pmatrix}.$

$\overrightarrow{AD}=\begin{pmatrix}12-0\\11-7\end{pmatrix}=\begin{pmatrix}12\\4\end{pmatrix}.$

Since
$\begin{pmatrix}12\\4\end{pmatrix}=2\begin{pmatrix}6\\2\end{pmatrix},$
therefore
$\overrightarrow{AD}=2\overrightarrow{AB}.$