The line l has equation $\frac{x-1}{-2}=y=\frac{z+7}{4}$, and the plane p has equation $x-z=2$.

1. Extract the direction vector of the line and the normal vector of the plane from their Cartesian equations.

From the line equation $\frac{x-1}{-2} = y = \frac{z+7}{4}$, the direction vector is:
$\mathbf{d} = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix}$

From the plane equation $x - z = 2$, the normal vector is:
$\mathbf{n} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$

2. Calculate the scalar product and the magnitudes of the two vectors.

$\mathbf{d} \cdot \mathbf{n} = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = (-2)(1) + (1)(0) + (4)(-1) = -6$

$|\mathbf{d}| = \sqrt{(-2)^2 + 1^2 + 4^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$

$|\mathbf{n}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}$

3. Use the sine formula to find the acute angle between the line and the plane.

Let $\theta$ be the acute angle between $l$ and $p$.
$\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$
$\sin \theta = \frac{|-6|}{\sqrt{21}\sqrt{2}} = \frac{6}{\sqrt{42}}$
$\theta = \sin^{-1} \left( \frac{6}{\sqrt{42}} \right) \approx 67.792^\circ$
$\theta = 67.8^\circ \text{ (to 1 d.p.)}$

1. Express the coordinates of a general point on the line $l$ in terms of a parameter $\lambda$.

Let $\frac{x-1}{-2} = y = \frac{z+7}{4} = \lambda$.
Then, the parametric equations of the line are:
$x = 1 - 2\lambda$
$y = \lambda$
$z = -7 + 4\lambda$

2. Substitute the parametric equations into the equation of the plane $p$ to find the value of $\lambda$ at the intersection point.

Substitute $x$ and $z$ into $x - z = 2$:
$(1 - 2\lambda) - (-7 + 4\lambda) = 2$
$1 - 2\lambda + 7 - 4\lambda = 2$
$8 - 6\lambda = 2$
$6\lambda = 6$
$\lambda = 1$

3. Calculate the exact coordinates using the value of $\lambda$.

Substitute $\lambda = 1$ back into the parametric equations:
$x = 1 - 2(1) = -1$
$y = 1$
$z = -7 + 4(1) = -3$

The coordinates of the intersection point are $(-1, 1, -3)$.

1. Find the normal vector to the plane and form the equation of the perpendicular line.

The equation of the plane $p$ is $x - z = 2$, so its normal vector is $\mathbf{n} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$.
Let the perpendicular line from $(1, 0, -7)$ to $p$ be $l_N$. Its equation is:
$\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ -7 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 + \lambda \\ 0 \\ -7 - \lambda \end{pmatrix}$

2. Substitute the line equation into the plane equation to find the value of the parameter.

Since $N$ lies on the plane $p$, its coordinates must satisfy $x - z = 2$. Substituting the components of the general point on $l_N$:
$(1 + \lambda) - (-7 - \lambda) = 2$
$8 + 2\lambda = 2$
$2\lambda = -6$
$\lambda = -3$

3. Determine the position vector of $N$.

Substitute $\lambda = -3$ back into the equation of the line $l_N$:
$\vec{ON} = \begin{pmatrix} 1 - 3 \\ 0 \\ -7 - (-3) \end{pmatrix} = \begin{pmatrix} -2 \\ 0 \\ -4 \end{pmatrix}$

1. Determine the point of intersection $I$ between the line $l$ and the plane $p$ (calculated in part ii).

To find the intersection of line $l$ and plane $p$, we express $l$ in parametric form: $\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ -7 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix}$.
Substituting $x = 1 - 2\lambda$, $y = \lambda$, and $z = -7 + 4\lambda$ into the plane equation $x - z = 2$:
$(1 - 2\lambda) - (-7 + 4\lambda) = 2$
$8 - 6\lambda = 2 \implies 6\lambda = 6 \implies \lambda = 1$
The position vector of the intersection point $I$ is:
$\vec{OI} = \begin{pmatrix} 1 - 2(1) \\ 0 + 1 \\ -7 + 4(1) \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ -3 \end{pmatrix}$

2. Determine the foot of the perpendicular $N$ from the point $A(1, 0, -7)$ on the line to the plane $p$ (calculated in part iii).

Let $A(1, 0, -7)$ be the fixed point on $l$. The normal vector of $p$ is $\mathbf{n} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$.
The line passing through $A$ and perpendicular to $p$ has the equation $\mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ -7 \end{pmatrix} + t \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$.
Substituting into $x - z = 2$ to find the foot of the perpendicular $N$:
$(1 + t) - (-7 - t) = 2$
$8 + 2t = 2 \implies 2t = -6 \implies t = -3$
The position vector of $N$ is:
$\vec{ON} = \begin{pmatrix} 1 - 3 \\ 0 \\ -7 - (-3) \end{pmatrix} = \begin{pmatrix} -2 \\ 0 \\ -4 \end{pmatrix}$

3. Find the position vector of the reflected point $A'$ of $A$ in the plane $p$.

Let $A'$ be the reflection of $A(1, 0, -7)$ in the plane $p$. Since the foot of the perpendicular $N$ is the midpoint of the line segment $AA'$:
$\vec{OA'} = 2\vec{ON} - \vec{OA}$
$\vec{OA'} = 2\begin{pmatrix} -2 \\ 0 \\ -4 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ -7 \end{pmatrix} = \begin{pmatrix} -4 \\ 0 \\ -8 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ -7 \end{pmatrix} = \begin{pmatrix} -5 \\ 0 \\ -1 \end{pmatrix}$

4. Formulate the vector equation of the reflected line passing through $I$ and $A'$.

The reflected line must pass through the point of intersection $I(-1, 1, -3)$ and the reflected point $A'(-5, 0, -1)$.
The direction vector of the reflected line is:
$\vec{IA'} = \vec{OA'} - \vec{OI} = \begin{pmatrix} -5 \\ 0 \\ -1 \end{pmatrix} - \begin{pmatrix} -1 \\ 1 \\ -3 \end{pmatrix} = \begin{pmatrix} -4 \\ -1 \\ 2 \end{pmatrix}$
Hence, using point $I$, the vector equation of the reflected line is:
$\mathbf{r} = \begin{pmatrix} -1 \\ 1 \\ -3 \end{pmatrix} + \mu \begin{pmatrix} -4 \\ -1 \\ 2 \end{pmatrix}$