The line $l_1$ and the plane $\pi_1$ have equations $\mathbf{r} = \begin{pmatrix} 1 \\ -3 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}, \lambda \in \mathbb{R}$ and $\mathbf{r} \bullet \begin{pmatrix} 2 \\ -5 \\ p \end{pmatrix} = 4$ respectively, where $p$ is a real constant.

1. Use the condition that the point of intersection lies on the plane $\pi_1$.

Since the point $A(-5, -7, 7)$ is the intersection of $l_1$ and $\pi_1$, it must lie on the plane $\pi_1$. Therefore, its position vector satisfies the plane's equation:
$\begin{pmatrix} -5 \\ -7 \\ 7 \end{pmatrix} \bullet \begin{pmatrix} 2 \\ -5 \\ p \end{pmatrix} = 4$

2. Evaluate the scalar product and solve for $p$.

$(-5)(2) + (-7)(-5) + 7(p) = 4$
$-10 + 35 + 7p = 4$
$25 + 7p = 4$
$7p = 4 - 25$
$7p = -21$
$p = -3 \text{ (shown)}$

1. Identify the direction vector of the line and the normal vector of the plane, and compute their scalar product and magnitudes.

From part (i), $p = -3$.
The direction vector of $l_1$ is $\mathbf{d} = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix}$ and the normal vector of $\pi_1$ is $\mathbf{n} = \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix}$.

$\mathbf{d} \bullet \mathbf{n} = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \bullet \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = (3)(2) + (2)(-5) + (-2)(-3) = 6 - 10 + 6 = 2$

$|\mathbf{d}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{17}$
$|\mathbf{n}| = \sqrt{2^2 + (-5)^2 + (-3)^2} = \sqrt{38}$

2. Use the sine formula for the acute angle between a line and a plane.

Let $\theta$ be the acute angle between $l_1$ and $\pi_1$. The formula is given by $\sin \theta = \frac{|\mathbf{d} \bullet \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}$.

$\sin \theta = \frac{|2|}{\sqrt{17}\sqrt{38}} = \frac{2}{\sqrt{646}}$

$\theta = \sin^{-1} \left( \frac{2}{\sqrt{646}} \right)$
$\theta \approx 4.5126^\circ \approx 4.5^\circ \text{ (to 1 d.p.)}$

1. Find the position vector of point $B$.

Substitute $\lambda = 1$ into the equation of $l_1$:
$\vec{OB} = \begin{pmatrix} 1 \\ -3 \\ 3 \end{pmatrix} + (1) \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix}$

2. Form the equation of the line passing through $B$ and perpendicular to $\pi_1$.

From part (i), $p = -3$, so the normal vector to $\pi_1$ is $\mathbf{n} = \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix}$.
The line passing through $B$ and perpendicular to $\pi_1$, let's call it $l_{BF}$, has equation:
$\mathbf{r} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix}, \quad \mu \in \mathbb{R}$

3. Substitute the line equation into the plane equation to find the value of $\mu$ at the foot of perpendicular $F$.

Since $F$ lies on $\pi_1$, substitute its position vector into the equation of $\pi_1$:
$\left[ \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} \right] \bullet \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = 4$
$\begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} \bullet \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} \bullet \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = 4$
$(8 + 5 - 3) + \mu(4 + 25 + 9) = 4$
$10 + 38\mu = 4 \implies 38\mu = -6 \implies \mu = -\frac{3}{19}$

4. Calculate the position vector of $F$.

Substitute $\mu = -\frac{3}{19}$ back into the equation of $l_{BF}$:
$\vec{OF} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} - \frac{3}{19} \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix}$
$\vec{OF} = \begin{pmatrix} \frac{76}{19} - \frac{6}{19} \\ -\frac{19}{19} + \frac{15}{19} \\ \frac{19}{19} + \frac{9}{19} \end{pmatrix} = \begin{pmatrix} \frac{70}{19} \\ -\frac{4}{19} \\ \frac{28}{19} \end{pmatrix}$

1. Work out the position vector of $F$, the foot of the perpendicular from $B$ to $\pi_1$ (required from part iii).

From part (iii), substitute $\lambda = 1$ to find $\overrightarrow{OB}$:
$\overrightarrow{OB} = \begin{pmatrix} 1 \\ -3 \\ 3 \end{pmatrix} + (1)\begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix}$
The line passing through $B$ perpendicular to $\pi_1$ is $\mathbf{r} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix}$.
Substituting this into the equation of $\pi_1$:
$\begin{pmatrix} 4 + 2\mu \\ -1 - 5\mu \\ 1 - 3\mu \end{pmatrix} \bullet \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = 4$
$2(4 + 2\mu) - 5(-1 - 5\mu) - 3(1 - 3\mu) = 4$
$10 + 38\mu = 4 \implies \mu = -\frac{3}{19}$
$\overrightarrow{OF} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} - \frac{3}{19} \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix} = \frac{1}{19} \begin{pmatrix} 70 \\ -4 \\ 28 \end{pmatrix}$

2. Find the position vector of $B'$, the reflection of $B$ in the plane $\pi_1$.

Let $B'$ be the reflection of $B$ in $\pi_1$. The foot of the perpendicular $F$ is the midpoint of $B$ and $B'$:
$\overrightarrow{OF} = \frac{\overrightarrow{OB} + \overrightarrow{OB'}}{2} \implies \overrightarrow{OB'} = 2\overrightarrow{OF} - \overrightarrow{OB}$
$\overrightarrow{OB'} = \frac{2}{19} \begin{pmatrix} 70 \\ -4 \\ 28 \end{pmatrix} - \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} = \frac{1}{19} \begin{pmatrix} 140 - 76 \\ -8 + 19 \\ 56 - 19 \end{pmatrix} = \frac{1}{19} \begin{pmatrix} 64 \\ 11 \\ 37 \end{pmatrix}$

3. Find the direction vector of the reflected line using points $A$ and $B'$ and state its equation.

The reflected line passes through the intersection point $A(-5, -7, 7)$ (from part i) and the reflected point $B'$.
Its direction vector is $\overrightarrow{AB'}$:
$\overrightarrow{AB'} = \overrightarrow{OB'} - \overrightarrow{OA} = \frac{1}{19} \begin{pmatrix} 64 \\ 11 \\ 37 \end{pmatrix} - \begin{pmatrix} -5 \\ -7 \\ 7 \end{pmatrix} = \frac{1}{19} \begin{pmatrix} 64 + 95 \\ 11 + 133 \\ 37 - 133 \end{pmatrix} = \frac{1}{19} \begin{pmatrix} 159 \\ 144 \\ -96 \end{pmatrix}$
Factoring out the scalar $\frac{3}{19}$, a parallel direction vector is $\begin{pmatrix} 53 \\ 48 \\ -32 \end{pmatrix}$.
Therefore, using point $A$, the equation of the reflected line is:
$\mathbf{r} = \begin{pmatrix} -5 \\ -7 \\ 7 \end{pmatrix} + t \begin{pmatrix} 53 \\ 48 \\ -32 \end{pmatrix}, t \in \mathbb{R}$

1. Find the position vector of point $B$.

Substitute $\lambda = 1$ into the equation of $l_1$:
$\vec{OB} = \begin{pmatrix} 1 \\ -3 \\ 3 \end{pmatrix} + (1) \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix}$

2. Find the direction vector of $l_2$.

Since $l_2$ is perpendicular to $l_1$ and parallel to $\pi_1$, its direction vector $\mathbf{d}_2$ must be perpendicular to both the direction vector of $l_1$ and the normal vector of $\pi_1$. We find $\mathbf{d}_2$ using the cross product:
$\mathbf{d}_2 = \begin{pmatrix} 3 \\ 2 \\ -2 \end{pmatrix} \times \begin{pmatrix} 2 \\ -5 \\ -3 \end{pmatrix}$
$= \begin{pmatrix} (2)(-3) - (-2)(-5) \\ (-2)(2) - (3)(-3) \\ (3)(-5) - (2)(2) \end{pmatrix}$
$= \begin{pmatrix} -16 \\ 5 \\ -19 \end{pmatrix}$

3. Write the vector equation of $l_2$.

Using the position vector of $B$ and the direction vector $\mathbf{d}_2$:
$l_2: \mathbf{r} = \begin{pmatrix} 4 \\ -1 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} -16 \\ 5 \\ -19 \end{pmatrix}, \mu \in \mathbb{R}$