Referred to the origin $O$, the points $P$ and $Q$ have position vectors $4\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}$ and $8\mathbf{i} + 8\mathbf{j} - \mathbf{k}$ respectively.

1. Find the direction vector $\overrightarrow{PQ}$.

$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \begin{pmatrix} 8 \\ 8 \\ -1 \end{pmatrix} - \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \\ -4 \end{pmatrix}$

2. Formulate the vector equation of the line $PQ$ and use the property of the $x$-$y$ plane to find the parameter.

The equation of line $PQ$ is $\mathbf{r} = \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 4 \\ -4 \end{pmatrix}$, $\lambda \in \mathbb{R}$.
Since $R$ lies on the $x$-$y$ plane, its $z$-coordinate is $0$.
$3 - 4\lambda = 0 \implies \lambda = \frac{3}{4}$

3. Substitute $\lambda$ to find the position vector of $R$.

$\overrightarrow{OR} = \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} + \frac{3}{4} \begin{pmatrix} 4 \\ 4 \\ -4 \end{pmatrix} = \begin{pmatrix} 7 \\ 7 \\ 0 \end{pmatrix}$

The position vector of $R$ is $7\mathbf{i} + 7\mathbf{j}$.

1. Find the position vector of $R$ (required from part (i)).

The line $PQ$ has direction vector $\vec{PQ} = \vec{OQ} - \vec{OP} = \begin{pmatrix} 8 \\ 8 \\ -1 \end{pmatrix} - \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \\ -4 \end{pmatrix}$.
The equation of line $PQ$ is $\mathbf{r} = \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 4 \\ -4 \end{pmatrix}$.
Since $R$ lies on the $x$-$y$ plane, its $z$-coordinate is $0$:
$3 - 4\lambda = 0 \implies \lambda = \frac{3}{4}$
$\vec{OR} = \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} + \frac{3}{4} \begin{pmatrix} 4 \\ 4 \\ -4 \end{pmatrix} = \begin{pmatrix} 7 \\ 7 \\ 0 \end{pmatrix}$

2. Identify the normal vector to plane $\Pi$ and determine its cartesian equation.

Since the plane $\Pi$ is perpendicular to $PQ$, its normal vector $\mathbf{n}$ is parallel to $\vec{PQ}$.
Taking a simplified normal vector $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$, the scalar product equation of the plane $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$ is:
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 7 \\ 7 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$
$x + y - z = 7(1) + 7(1) + 0(-1)$
$x + y - z = 14$
Thus, $c = 14$.

1. Identify the normal vectors of plane $\Pi$ and the $x$-$y$ plane

From the cartesian equation $x + y - z = c$ given in part (ii), the normal vector to plane $\Pi$ is $\mathbf{n}_1 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.

The normal vector to the $x$-$y$ plane is parallel to the $z$-axis, so $\mathbf{n}_2 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

2. Calculate the acute angle between the two planes using the dot product

Let $\theta$ be the acute angle between the planes. The angle between the planes is the acute angle between their normal vectors.

$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$

$\cos \theta = \frac{|(1)(0) + (1)(0) + (-1)(1)|}{\sqrt{1^2 + 1^2 + (-1)^2} \sqrt{0^2 + 0^2 + 1^2}}$

$\cos \theta = \frac{|-1|}{\sqrt{3}(1)} = \frac{1}{\sqrt{3}}$

$\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right) \approx 54.735...^\circ$

1. Find the position vector of $R$ and the vector $\overrightarrow{AR}$.

The direction of $PQ$ is $\overrightarrow{PQ} = \begin{pmatrix} 8 \\ 8 \\ -1 \end{pmatrix} - \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \\ -4 \end{pmatrix}$.
The line $PQ$ has the equation $\mathbf{r} = \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.
The $x$-$y$ plane has the equation $z=0$. Substituting this into the line equation gives $3 - \lambda = 0 \implies \lambda = 3$.
The position vector of $R$ is $\overrightarrow{OR} = \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix} + 3\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 7 \\ 7 \\ 0 \end{pmatrix}$.
Given $\overrightarrow{OA} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$, we have:
$\overrightarrow{AR} = \overrightarrow{OR} - \overrightarrow{OA} = \begin{pmatrix} 7 \\ 7 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 6 \\ 6 \\ 0 \end{pmatrix}$.

2. Find the direction vector of the line of intersection of $\Pi$ and the $x$-$y$ plane.

The plane $\Pi$ is perpendicular to $PQ$, so its normal vector is $\mathbf{n}_1 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.
The normal to the $x$-$y$ plane is $\mathbf{n}_2 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.
The direction vector $\mathbf{d}$ of their line of intersection is mutually perpendicular to both normal vectors. Using the cross product:
$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} (1)(1) - (-1)(0) \\ (-1)(0) - (1)(1) \\ (1)(0) - (1)(0) \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$.

3. Use the dot product to show that $\overrightarrow{AR}$ is perpendicular to the line of intersection.

$\overrightarrow{AR} \cdot \mathbf{d} = \begin{pmatrix} 6 \\ 6 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} = 6(1) + 6(-1) + 0(0) = 0$.
Since the dot product is $0$, $\overrightarrow{AR}$ is perpendicular to the line of intersection of $\Pi$ and the $x$-$y$ plane.

1. Determine the value of $c$ for the cartesian equation of plane $\Pi$ from earlier parts.

From part (i), the position vector of $R$ is $7\mathbf{i} + 7\mathbf{j}$.
Since $R$ lies on the plane $\Pi$ with equation $x + y - z = c$, we substitute the coordinates of $R(7, 7, 0)$ into the equation:
$7 + 7 - 0 = c \implies c = 14$
Thus, the equation of $\Pi$ is $x + y - z = 14$.

2. Apply the exact distance formula from a point to a plane.

The shortest distance from a point $A(x_0, y_0, z_0)$ to a plane $ax + by + cz = d$ is given by $D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$.
Using $A(1, 1, 0)$ and $\Pi: x + y - z = 14$,
$D = \frac{|(1)(1) + (1)(1) + (-1)(0) - 14|}{\sqrt{1^2 + 1^2 + (-1)^2}}$

3. Evaluate and simplify the distance expression to an exact surd.

$D = \frac{|2 - 14|}{\sqrt{3}}$
$D = \frac{|-12|}{\sqrt{3}}$
$D = \frac{12}{\sqrt{3}} = 4\sqrt{3}$