Relative to the origin $O$, the point $A$ has position vector $3\mathbf{i} + 7\mathbf{j} + \mathbf{k}$. The plane $\pi$ contains $A$ and is parallel to the vector $-2\mathbf{i} + 2\mathbf{j} + \mathbf{k}$ and the line $\mathbf{r} = \alpha(-\mathbf{i} + 2\mathbf{j} + 3\mathbf{k})$, where $\alpha \in \mathbb{R}$.

1. Find the normal vector to the plane by taking the cross product of the two direction vectors parallel to the plane.

$\mathbf{n} = \begin{pmatrix} -2 \\ 2 \\ 1 \end{pmatrix} \times \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}$
$\mathbf{n} = \begin{pmatrix} (2)(3) - (1)(2) \\ (1)(-1) - (-2)(3) \\ (-2)(2) - (2)(-1) \end{pmatrix} = \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix}$

2. Substitute the position vector of point $A$ to evaluate the scalar product $\mathbf{a} \cdot \mathbf{n}$.

$\mathbf{a} = \begin{pmatrix} 3 \\ 7 \\ 1 \end{pmatrix}$
$\mathbf{a} \cdot \mathbf{n} = \begin{pmatrix} 3 \\ 7 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix}$
$\mathbf{a} \cdot \mathbf{n} = 3(4) + 7(5) + 1(-2) = 12 + 35 - 2 = 45$

3. Formulate the cartesian equation of the plane.

The vector equation of the plane is $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$.
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix} = 45$
$4x + 5y - 2z = 45 \quad \text{(Shown)}$

1. Formulate the vector equation of the line passing through $B$ and perpendicular to the plane $\pi$.

Let $l_{BF}$ be the line passing through $B$ and perpendicular to $\pi$.
From the cartesian equation of $\pi$, $4x + 5y - 2z = 45$, the normal vector to the plane is $\mathbf{n} = \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix}$. This acts as the direction vector of $l_{BF}$.
The vector equation of $l_{BF}$ is:
$\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix}$, where $\lambda \in \mathbb{R}$.

2. Substitute the line equation into the plane equation to find the parameter $\lambda$ at the point of intersection.

Since $F$ lies on $\pi$, its position vector satisfies the plane's scalar product equation $\mathbf{r} \cdot \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix} = 45$.
Substituting the equation of $l_{BF}$ into the plane equation:
$\left[ \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix} \right] \cdot \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix} = 45$
$\begin{pmatrix} 2+4\lambda \\ 5\lambda \\ 4-2\lambda \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix} = 45$
$4(2+4\lambda) + 5(5\lambda) - 2(4-2\lambda) = 45$
$8 + 16\lambda + 25\lambda - 8 + 4\lambda = 45$
$45\lambda = 45$
$\lambda = 1$

3. Calculate the position vector of $F$.

Substitute $\lambda = 1$ back into the vector equation of $l_{BF}$:
$\overrightarrow{OF} = \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix} + (1)\begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 6 \\ 5 \\ 2 \end{pmatrix}$
Expressed in $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ notation, the position vector of $F$ is $6\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}$.

1. Calculate the position vector of $F$, the foot of the perpendicular from $B$ to the plane (required result from part ii).

The line passing through $B$ perpendicular to $\pi$ has a direction vector equal to the normal of $\pi$, $\mathbf{n} = \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix}$.
The equation of this line is $\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix}$.
Substitute this into the plane equation $4x + 5y - 2z = 45$:
$4(2 + 4\lambda) + 5(5\lambda) - 2(4 - 2\lambda) = 45$
$8 + 16\lambda + 25\lambda - 8 + 4\lambda = 45 \implies 45\lambda = 45 \implies \lambda = 1$
$\vec{OF} = \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix} + 1 \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix} = \begin{pmatrix} 6 \\ 5 \\ 2 \end{pmatrix}$

2. Use the foot of the perpendicular $F$ as the midpoint of $B$ and its reflection $B'$ to find the position vector of $B'$.

Since $B'$ is the reflection of $B$ in the plane $\pi$, $F$ is the midpoint of the line segment $BB'$.
$\vec{OF} = \frac{\vec{OB} + \vec{OB'}}{2}$
$\vec{OB'} = 2\vec{OF} - \vec{OB}$
$\vec{OB'} = 2 \begin{pmatrix} 6 \\ 5 \\ 2 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix}$
$\vec{OB'} = \begin{pmatrix} 12 \\ 10 \\ 4 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix} = \begin{pmatrix} 10 \\ 10 \\ 0 \end{pmatrix}$

1. Find the position vectors of $F$ (from part ii) and $B'$ (from part iii)

The line through $B$ perpendicular to $\pi$ is $\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 5 \\ -2 \end{pmatrix}$.
Substituting into the plane equation $4x + 5y - 2z = 45$:
$4(2+4\lambda) + 5(5\lambda) - 2(4-2\lambda) = 45 \implies 45\lambda = 45 \implies \lambda = 1$
So, $\vec{OF} = \begin{pmatrix} 6 \\ 5 \\ 2 \end{pmatrix}$.
Since $B'$ is the reflection of $B$ in $\pi$, $F$ is the midpoint of $BB'$:
$\vec{OB'} = 2\vec{OF} - \vec{OB} = 2\begin{pmatrix} 6 \\ 5 \\ 2 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix} = \begin{pmatrix} 10 \\ 10 \\ 0 \end{pmatrix}$

2. Find vectors $\vec{AB}$ and $\vec{AB'}$

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 2 \\ 0 \\ 4 \end{pmatrix} - \begin{pmatrix} 3 \\ 7 \\ 1 \end{pmatrix} = \begin{pmatrix} -1 \\ -7 \\ 3 \end{pmatrix}$
$\vec{AB'} = \vec{OB'} - \vec{OA} = \begin{pmatrix} 10 \\ 10 \\ 0 \end{pmatrix} - \begin{pmatrix} 3 \\ 7 \\ 1 \end{pmatrix} = \begin{pmatrix} 7 \\ 3 \\ -1 \end{pmatrix}$

3. Calculate the angle $BAB'$ using the scalar product

$\cos \angle BAB' = \frac{\vec{AB} \cdot \vec{AB'}}{|\vec{AB}| |\vec{AB'}|}$
$\vec{AB} \cdot \vec{AB'} = (-1)(7) + (-7)(3) + (3)(-1) = -7 - 21 - 3 = -31$
$|\vec{AB}| = \sqrt{(-1)^2 + (-7)^2 + 3^2} = \sqrt{1 + 49 + 9} = \sqrt{59}$
$|\vec{AB'}| = \sqrt{7^2 + 3^2 + (-1)^2} = \sqrt{49 + 9 + 1} = \sqrt{59}$
$\cos \angle BAB' = \frac{-31}{\sqrt{59} \sqrt{59}} = -\frac{31}{59}$
$\angle BAB' = \cos^{-1}\left(-\frac{31}{59}\right) \approx 121.69^\circ$