The points $P$ and $Q$ have position vectors $3\mathbf{i}+4\mathbf{j}-\mathbf{k}$ and $5\mathbf{i}+7\mathbf{j}+6\mathbf{k}$ respectively.

1. State a suitable position vector $\mathbf{a}$ for the plane.

Since the plane passes through point $Q$, its position vector can be taken as $\mathbf{a}$.
$\mathbf{a} = \vec{OQ} = \begin{pmatrix} 5 \\ 7 \\ 6 \end{pmatrix}$

2. Find the normal vector to the plane, $\vec{PQ}$.

$\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 5 \\ 7 \\ 6 \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix}$

3. Explain why $3\mathbf{i}-2\mathbf{j}$ can be taken as $\mathbf{b}$.

For a direction vector $\mathbf{b}$ to be parallel to the plane, it must be perpendicular to the normal vector $\vec{PQ}$. We test this using the scalar product:
$\mathbf{b} \cdot \vec{PQ} = \begin{pmatrix} 3 \\ -2 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix} = (3)(2) + (-2)(3) + (0)(7) = 6 - 6 = 0$
Since the scalar product is $0$, $\mathbf{b}$ is perpendicular to the normal $\vec{PQ}$ and hence is parallel to the plane $\Pi$. Therefore, it can be taken as a valid direction vector for the plane.

4. Find a suitable vector $\mathbf{c}$.

Vector $\mathbf{c}$ must be parallel to the plane (thus perpendicular to $\vec{PQ}$) and perpendicular to $\mathbf{b}$. A suitable vector $\mathbf{c}$ can be found using the vector product of the normal vector and $\mathbf{b}$.
$\mathbf{c} = \vec{PQ} \times \mathbf{b} = \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix} \times \begin{pmatrix} 3 \\ -2 \\ 0 \end{pmatrix}$
$\mathbf{c} = \begin{pmatrix} (3)(0) - (7)(-2) \\ (7)(3) - (2)(0) \\ (2)(-2) - (3)(3) \end{pmatrix} = \begin{pmatrix} 14 \\ 21 \\ -13 \end{pmatrix}$

1. Establish that both lines are coplanar by showing they both lie on the plane $\Pi$.

The point $Q$ (with position vector $\mathbf{a}$) and the point $S$ (with position vector $\mathbf{s}$) both lie on the plane $\Pi$. The given direction vectors $\mathbf{b}$ and $\mathbf{c}$ are both parallel to $\Pi$ as they form the equation of the plane $\Pi$. Therefore, the lines $\mathbf{r}=\mathbf{a}+\alpha\mathbf{b}$ and $\mathbf{r}=\mathbf{s}+\beta\mathbf{c}$ both lie completely in the same plane $\Pi$, meaning they are coplanar.

2. Use the relationship between the direction vectors to conclude that the lines must intersect.

From part (i), the vectors $\mathbf{b}$ and $\mathbf{c}$ are perpendicular to each other, which implies they are not parallel. Since the two lines are coplanar and not parallel, they cannot be skew and must therefore intersect at exactly one point.

1. Find the normal vector and the scalar product equation of the plane $\Pi$.

$\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 5 \\ 7 \\ 6 \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix}$
Since $\Pi$ is perpendicular to $PQ$, $\vec{PQ}$ is the normal vector to the plane. The equation of plane $\Pi$ is $\mathbf{r} \cdot \vec{PQ} = \vec{OQ} \cdot \vec{PQ}$:
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix} = \begin{pmatrix} 5 \\ 7 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix} = 5(2) + 7(3) + 6(7) = 73$

2. Find the vector equation of the line $l$ passing through $P$ and $R$.

$\vec{PR} = \vec{OR} - \vec{OP} = \begin{pmatrix} 6 \\ 43 \\ 8 \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 39 \\ 9 \end{pmatrix}$
We can take the direction vector of the line as $\mathbf{d} = \frac{1}{3}\vec{PR} = \begin{pmatrix} 1 \\ 13 \\ 3 \end{pmatrix}$.
The vector equation of line $l$ is:
$\mathbf{r} = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 13 \\ 3 \end{pmatrix}, \quad t \in \mathbb{R}$

3. Find the position vector of intersection point $S$.

Since $S$ lies on both $l$ and $\Pi$, substitute the line equation into the plane equation:
$\begin{pmatrix} 3 + t \\ 4 + 13t \\ -1 + 3t \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix} = 73$
$2(3 + t) + 3(4 + 13t) + 7(-1 + 3t) = 73$
$6 + 2t + 12 + 39t - 7 + 21t = 73$
$62t + 11 = 73$
$62t = 62 \implies t = 1$
Substitute $t = 1$ into the line equation:
$\vec{OS} = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} + (1)\begin{pmatrix} 1 \\ 13 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 17 \\ 2 \end{pmatrix}$
So the position vector of $S$ is $\mathbf{s} = 4\mathbf{i} + 17\mathbf{j} + 2\mathbf{k}$.

4. Determine whether $S$ lies on $PR$ produced.

Find the vector $\vec{PS}$:
$\vec{PS} = \vec{OS} - \vec{OP} = \begin{pmatrix} 4 \\ 17 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 13 \\ 3 \end{pmatrix}$
We have $\vec{PR} = \begin{pmatrix} 3 \\ 39 \\ 9 \end{pmatrix}$, so $\vec{PS} = \frac{1}{3}\vec{PR}$.
Since $\vec{PS}$ is in the same direction as $\vec{PR}$ and its magnitude is exactly $\frac{1}{3}$ that of $\vec{PR}$ ($0 < \frac{1}{3} < 1$), the point $S$ lies strictly between $P$ and $R$.
Therefore, $S$ does not lie on $PR$ produced.