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 $\Pi$.

Since the plane $\Pi$ passes through $Q$, $\mathbf{a}$ can be taken as the position vector of $Q$.
$\mathbf{a} = \vec{OQ} = 5\mathbf{i}+7\mathbf{j}+6\mathbf{k}$

2. Find the normal vector to the plane.

The plane is perpendicular to $PQ$, so the normal vector $\mathbf{n}$ is $\vec{PQ}$.
$\mathbf{n} = \vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 5 \\ 7 \\ 6 \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix}$

3. Show that $3\mathbf{i}-2\mathbf{j}$ is a valid direction vector for the plane.

Vector $\mathbf{b}$ must be a direction vector parallel to the plane, which means it must be perpendicular to the normal $\vec{PQ}$. We test this using the dot 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 dot product is $0$, the vector $3\mathbf{i}-2\mathbf{j}$ is perpendicular to the normal. Therefore, it is parallel to the plane and can be taken as $\mathbf{b}$.

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

Vector $\mathbf{c}$ must be parallel to the plane (hence perpendicular to $\vec{PQ}$) and also perpendicular to $\mathbf{b}$. We can find $\mathbf{c}$ by taking the cross product of the normal $\vec{PQ}$ and $\mathbf{b}$:
$\mathbf{c} = \vec{PQ} \times \mathbf{b} = \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix} \times \begin{pmatrix} 3 \\ -2 \\ 0 \end{pmatrix} = \begin{pmatrix} (3)(0) - (7)(-2) \\ (7)(3) - (2)(0) \\ (2)(-2) - (3)(3) \end{pmatrix} = \begin{pmatrix} 14 \\ 21 \\ -13 \end{pmatrix}$
$\mathbf{c} = 14\mathbf{i} + 21\mathbf{j} - 13\mathbf{k}$

1. Establish that both lines lie on the same plane (they are coplanar).

From part (i), the equation of the plane $\Pi$ is $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}$.
The line $\mathbf{r}=\mathbf{a}+\alpha\mathbf{b}$ corresponds to points on $\Pi$ where $\mu = 0$, so it lies entirely on $\Pi$.
Point $S$ is the intersection of line $l$ and $\Pi$, so $S$ lies on $\Pi$. The line $\mathbf{r}=\mathbf{s}+\beta\mathbf{c}$ passes through $S$ and is parallel to the direction vector $\mathbf{c}$ of the plane $\Pi$, so it also lies entirely on $\Pi$.
Therefore, both lines are coplanar.

2. Conclude that the lines intersect because they are not parallel.

From part (i), the direction vectors $\mathbf{b}$ and $\mathbf{c}$ are perpendicular to each other ($\mathbf{b} \cdot \mathbf{c} = 0$), which means the lines are not parallel.
In three-dimensional space, any two coplanar lines that are not parallel must intersect.

1. Find the normal vector to the plane $\Pi$ and form its scalar product equation.

The normal vector to the plane $\Pi$ is $\vec{PQ}$.
$\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 5 \\ 7 \\ 6 \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix}$
The equation of $\Pi$ passing through $Q$ is $\mathbf{r} \cdot \mathbf{n} = \vec{OQ} \cdot \mathbf{n}$:
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix} = \begin{pmatrix} 5 \\ 7 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix} = 10 + 21 + 42 = 73$

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

The direction vector of line $l$ is $\vec{PR}$.
$\vec{PR} = \vec{OR} - \vec{OP} = \begin{pmatrix} 6 \\ 43 \\ 8 \end{pmatrix} - \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 39 \\ 9 \end{pmatrix}$
The vector equation of line $l$ is:
$\mathbf{r} = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} + \gamma \begin{pmatrix} 3 \\ 39 \\ 9 \end{pmatrix}, \quad \gamma \in \mathbb{R}$

3. Substitute the line equation into the plane equation to find the value of the parameter $\gamma$ at intersection $S$.

Since point $S$ lies on both line $l$ and plane $\Pi$, we substitute the line's expression into the plane's equation:
$\left[ \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} + \gamma \begin{pmatrix} 3 \\ 39 \\ 9 \end{pmatrix} \right] \cdot \begin{pmatrix} 2 \\ 3 \\ 7 \end{pmatrix} = 73$
$(6 + 12 - 7) + \gamma(6 + 117 + 63) = 73$
$11 + 186\gamma = 73$
$186\gamma = 62 \implies \gamma = \frac{1}{3}$

4. Calculate the position vector of $S$.

Substitute $\gamma = \frac{1}{3}$ back into the equation of line $l$:
$\mathbf{s} = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} + \frac{1}{3} \begin{pmatrix} 3 \\ 39 \\ 9 \end{pmatrix}$
$\mathbf{s} = \begin{pmatrix} 3 \\ 4 \\ -1 \end{pmatrix} + \begin{pmatrix} 1 \\ 13 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 17 \\ 2 \end{pmatrix}$
Therefore, the position vector of point $S$ is $4\mathbf{i} + 17\mathbf{j} + 2\mathbf{k}$.

5. Determine whether $S$ lies on $PR$ produced by analysing the parameter value.

The position of a point on the line relative to $P$ and $R$ is given by the parameter $\gamma$ in the formulation $\vec{OS} = \vec{OP} + \gamma \vec{PR}$.

• $\gamma = 0$ corresponds to point $P$.
• $\gamma = 1$ corresponds to point $R$.
• $0 < \gamma < 1$ implies the point lies on the line segment strictly between $P$ and $R$.
• $\gamma > 1$ implies the point lies on the line $PR$ produced (beyond $R$).

Since $\gamma = \frac{1}{3}$, which satisfies $0 < \gamma < 1$, the point $S$ lies between $P$ and $R$.
Thus, point $S$ does not lie on $PR$ produced.