A home interior designer is looking to design a pull-down wardrobe lift. To test his concept, the designer built a small-scale model of it, in which points $(x, y, z)$ are defined relative to a corner of the model at $O(0, 0, 0)$. The rods used in the model can be modelled by equations of straight lines. The main rod can be modelled by the line $l_1$ with equation $\mathbf{r} = \begin{pmatrix} 5 \\ -5 \\ 8 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \lambda \in \mathbb{R}$. Another rod, line $l_2$, passes through the points $A(5, -5, 8)$ and $B(8, -4, 3)$.

1. Express the position vector of $F$ in terms of a parameter and find the displacement vector $\overrightarrow{BF}$.

Since $F$ lies on $l_1$, let its position vector be $\overrightarrow{OF} = \begin{pmatrix} 5+\lambda \\ -5+\lambda \\ 8 \end{pmatrix}$.

The displacement vector $\overrightarrow{BF}$ is:
$\overrightarrow{BF} = \overrightarrow{OF} - \overrightarrow{OB} = \begin{pmatrix} 5+\lambda \\ -5+\lambda \\ 8 \end{pmatrix} - \begin{pmatrix} 8 \\ -4 \\ 3 \end{pmatrix} = \begin{pmatrix} \lambda-3 \\ \lambda-1 \\ 5 \end{pmatrix}$

2. Apply the condition that $\overrightarrow{BF}$ is perpendicular to $l_1$ to solve for $\lambda$.

Since $\overrightarrow{BF}$ is perpendicular to $l_1$, its scalar product with the direction vector of $l_1$ is $0$.
$\begin{pmatrix} \lambda-3 \\ \lambda-1 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 0$
$(\lambda - 3)(1) + (\lambda - 1)(1) + (5)(0) = 0$
$2\lambda - 4 = 0 \implies \lambda = 2$

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

Substitute $\lambda = 2$ back into the expression for $\overrightarrow{OF}$:
$\overrightarrow{OF} = \begin{pmatrix} 5+2 \\ -5+2 \\ 8 \end{pmatrix} = \begin{pmatrix} 7 \\ -3 \\ 8 \end{pmatrix}$

1. Find the position vector of $F$ on $l_1$ such that $\overrightarrow{BF}$ is perpendicular to $l_1$ (as required from part a).

Let $F$ be a point on $l_1$ with position vector $\overrightarrow{OF} = \begin{pmatrix} 5 + \lambda \\ -5 + \lambda \\ 8 \end{pmatrix}$.
$\overrightarrow{BF} = \overrightarrow{OF} - \overrightarrow{OB} = \begin{pmatrix} 5 + \lambda \\ -5 + \lambda \\ 8 \end{pmatrix} - \begin{pmatrix} 8 \\ -4 \\ 3 \end{pmatrix} = \begin{pmatrix} \lambda - 3 \\ \lambda - 1 \\ 5 \end{pmatrix}$
Since $\overrightarrow{BF}$ is perpendicular to $l_1$:
$\begin{pmatrix} \lambda - 3 \\ \lambda - 1 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = 0$
$(\lambda - 3)(1) + (\lambda - 1)(1) = 0 \implies 2\lambda = 4 \implies \lambda = 2$
$\overrightarrow{OF} = \begin{pmatrix} 5 + 2 \\ -5 + 2 \\ 8 \end{pmatrix} = \begin{pmatrix} 7 \\ -3 \\ 8 \end{pmatrix}$

2. Find the position vector of $B'$, the reflection of point $B$ about $l_1$.

The point $F$ is the midpoint of the line segment connecting $B$ and its reflection $B'$.
$\overrightarrow{OF} = \frac{\overrightarrow{OB} + \overrightarrow{OB'}}{2}$
$\overrightarrow{OB'} = 2\overrightarrow{OF} - \overrightarrow{OB} = 2\begin{pmatrix} 7 \\ -3 \\ 8 \end{pmatrix} - \begin{pmatrix} 8 \\ -4 \\ 3 \end{pmatrix}$
$\overrightarrow{OB'} = \begin{pmatrix} 14 \\ -6 \\ 16 \end{pmatrix} - \begin{pmatrix} 8 \\ -4 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \\ 13 \end{pmatrix}$

3. State the intersection point $A$ and find the vector equation of the reflected line $l_R$.

The point $A(5, -5, 8)$ matches the fixed point in the equation of $l_1$ (when $\lambda = 0$), meaning $A$ is the intersection point of $l_1$ and $l_2$. Therefore, $A$ reflects to itself and lies on $l_R$.
The direction vector of $l_R$ is given by:
$\overrightarrow{AB'} = \overrightarrow{OB'} - \overrightarrow{OA} = \begin{pmatrix} 6 \\ -2 \\ 13 \end{pmatrix} - \begin{pmatrix} 5 \\ -5 \\ 8 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 5 \end{pmatrix}$
The vector equation of $l_R$, passing through $A$ with direction $\overrightarrow{AB'}$, is:
$\mathbf{r} = \begin{pmatrix} 5 \\ -5 \\ 8 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 3 \\ 5 \end{pmatrix}, \quad \mu \in \mathbb{R}$

1. Find a second direction vector on the plane by finding the displacement vector between a point on $l_1$ and $C$.

Let $A(5, -5, 8)$ be a point on $l_1$.
$\overrightarrow{CA} = \begin{pmatrix} 5 \\ -5 \\ 8 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix} = \begin{pmatrix} 5 \\ -5 \\ 2 \end{pmatrix}$

2. Calculate the cross product of the direction vector of $l_1$ and $\overrightarrow{CA}$ to find a normal vector to the plane.

Let $\mathbf{d_1} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ be the direction vector of $l_1$.
$\mathbf{n} = \mathbf{d_1} \times \overrightarrow{CA} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 5 \\ -5 \\ 2 \end{pmatrix} = \begin{pmatrix} (1)(2) - (0)(-5) \\ (0)(5) - (1)(2) \\ (1)(-5) - (1)(5) \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ -10 \end{pmatrix}$
A simpler normal vector is $\mathbf{n}' = \begin{pmatrix} -1 \\ 1 \\ 5 \end{pmatrix}$.

3. Formulate the cartesian equation of the plane using the scalar product equation.

$\mathbf{r} \cdot \mathbf{n}' = \overrightarrow{OC} \cdot \mathbf{n}'$
$\mathbf{r} \cdot \begin{pmatrix} -1 \\ 1 \\ 5 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 1 \\ 5 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} -1 \\ 1 \\ 5 \end{pmatrix} = 0 + 0 + 30 = 30$
Thus, the cartesian equation is $-x + y + 5z = 30$ (Shown).

1. Find the direction vector of the line of intersection by taking the cross product of the normal vectors of the two planes.

The normal vector of $p_1$ is $\mathbf{n}_1 = \begin{pmatrix} -1 \\ 1 \\ 5 \end{pmatrix}$ and the normal vector of $p_2$ is $\mathbf{n}_2 = \begin{pmatrix} -17 \\ -37 \\ 4 \end{pmatrix}$.

The direction vector $\mathbf{d}$ of the line of intersection $l_S$ is perpendicular to both normal vectors:
$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} -1 \\ 1 \\ 5 \end{pmatrix} \times \begin{pmatrix} -17 \\ -37 \\ 4 \end{pmatrix} = \begin{pmatrix} (1)(4) - (5)(-37) \\ (5)(-17) - (-1)(4) \\ (-1)(-37) - (1)(-17) \end{pmatrix}$
$\mathbf{d} = \begin{pmatrix} 4 + 185 \\ -85 + 4 \\ 37 + 17 \end{pmatrix} = \begin{pmatrix} 189 \\ -81 \\ 54 \end{pmatrix}$
We can scale this vector by a factor of $\frac{1}{27}$ to obtain a simpler direction vector: $\begin{pmatrix} 7 \\ -3 \\ 2 \end{pmatrix}$.

2. Find a common point that lies on both planes.

We need to find a coordinate point $(x, y, z)$ that satisfies the equations of both planes:
$-x + y + 5z = 30$
$-17x - 37y + 4z = 24$
By letting $x = 0$, the equations simplify to:
$y + 5z = 30 \implies y = 30 - 5z$
Substitute into the second equation:
$-37(30 - 5z) + 4z = 24$
$-1110 + 185z + 4z = 24$
$189z = 1134 \implies z = 6$
Substituting $z = 6$ back gives $y = 30 - 5(6) = 0$.
So, a common point on both planes is $(0, 0, 6)$.
(Note: This is the attachment point $C(0, 0, 6)$ given in part (c), which can be checked directly against the equation of $p_2$.)

3. Write the vector equation of the line $l_S$.

Using the position vector of the common point and the direction vector, the vector equation of the structural rod is:
$\mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ 6 \end{pmatrix} + \mu \begin{pmatrix} 7 \\ -3 \\ 2 \end{pmatrix}, \mu \in \mathbb{R}$

1. Deduce the value of $\alpha$ by recognising that the planes must be parallel.

Since the distance between planes $p_1$ and $p_3$ is exactly $3\sqrt{3}$ units everywhere, the two planes must not intersect, meaning they are parallel. Thus, their normal vectors are scalar multiples of each other:
$\begin{pmatrix} -3 \\ \alpha \\ 15 \end{pmatrix} = k \begin{pmatrix} -1 \\ 1 \\ 5 \end{pmatrix}$
Comparing the $x$-components or $z$-components, we find $-3 = -k \implies k = 3$.
Therefore, comparing the $y$-components:
$\alpha = 3(1) = 3$

2. Formulate an equation for the distance between the parallel planes.

With $\alpha=3$, the equation of $p_3$ is $-3x + 3y + 15z = \beta$.
Dividing by 3 gives an equation with the exact same normal vector as $p_1$ from part (c):
$p_3: -x + y + 5z = \frac{\beta}{3}$
The distance between the parallel planes $p_1: -x + y + 5z = 30$ and $p_3$ is:
$\text{Distance} = \frac{\left| 30 - \frac{\beta}{3} \right|}{\sqrt{(-1)^2 + 1^2 + 5^2}} = \frac{\left| 30 - \frac{\beta}{3} \right|}{\sqrt{27}}$

3. Solve for the possible values of $\beta$.

Equating the distance to $3\sqrt{3}$ and using $\sqrt{27} = 3\sqrt{3}$:
$\frac{\left| 30 - \frac{\beta}{3} \right|}{3\sqrt{3}} = 3\sqrt{3}$
$\left| 30 - \frac{\beta}{3} \right| = 27$
$30 - \frac{\beta}{3} = 27 \quad \text{or} \quad 30 - \frac{\beta}{3} = -27$
$\frac{\beta}{3} = 3 \quad \text{or} \quad \frac{\beta}{3} = 57$
$\beta = 9 \quad \text{or} \quad \beta = 171$