直线 $l$ 的方程为 $1 - x = \frac{z - \alpha}{2}, y = 1$，其中 $\alpha$ 为常数。

1. Substitute the point $(0, 1, 5)$ into the equation of line $l$.

Since the point $(0, 1, 5)$ lies on the line $l$, its coordinates must satisfy the line's equation $1 - x = \frac{z - \alpha}{2}, y = 1$.
Substituting $x = 0$ and $z = 5$ gives:
$1 - 0 = \frac{5 - \alpha}{2}$

2. Solve for the unknown constant $\alpha$.

$1 = \frac{5 - \alpha}{2}$
$2 = 5 - \alpha$
$\alpha = 3 \text{ (shown)}$

1. Identify a point on line $l$ and its direction vector.

From part (i), we know a point on line $l$ is $C(0, 1, 5)$.
Rewrite the equation of line $l$: $1 - x = \frac{z - 3}{2}, y = 1$ in standard form as $\frac{x - 1}{-1} = \frac{y - 1}{0} = \frac{z - 3}{2}$.
The direction vector of line $l$ is $\mathbf{d} = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}$.

2. Find a second vector lying on the plane $\pi_1$.

The plane $\pi_1$ contains the point $A(0, -1, 3)$ and the line $l$ which has the point $C(0, 1, 5)$.
A vector connecting point $A$ to point $C$ on the line is:
$\vec{AC} = \vec{OC} - \vec{OA} = \begin{pmatrix} 0 \\ 1 \\ 5 \end{pmatrix} - \begin{pmatrix} 0 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}$

3. Calculate the normal vector to the plane.

The normal vector $\mathbf{n}$ is perpendicular to both the direction vector of the line and the vector $\vec{AC}$.
$\mathbf{n} = \mathbf{d} \times \vec{AC} = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} \times \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} (0)(2) - (2)(2) \\ (2)(0) - (-1)(2) \\ (-1)(2) - (0)(0) \end{pmatrix} = \begin{pmatrix} -4 \\ 2 \\ -2 \end{pmatrix}$
We can scale this vector to get a simpler parallel vector as the normal: $\mathbf{n_1} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$.

4. Form the cartesian equation of the plane.

Using the normal vector $\mathbf{n_1}$ and point $A$, the equation of the plane $\pi_1$ is $\mathbf{r} \cdot \mathbf{n_1} = \vec{OA} \cdot \mathbf{n_1}$.
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$
$2x - y + z = (0)(2) + (-1)(-1) + (3)(1)$
$2x - y + z = 4$

1. Find the direction vector of the line $AB$.

$\vec{AB} = B - A = \begin{pmatrix} 12 \\ 5 \\ -1 \end{pmatrix} - \begin{pmatrix} 0 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 12 \\ 6 \\ -4 \end{pmatrix} = 2\begin{pmatrix} 6 \\ 3 \\ -2 \end{pmatrix}$
We can take the direction vector of the line $AB$ as $\mathbf{m} = \begin{pmatrix} 6 \\ 3 \\ -2 \end{pmatrix}$.

2. Determine the normal vector to the plane $\pi_1$ (required from part (ii)).

From part (i), line $l$ has equation $1 - x = \frac{z - 3}{2}, y = 1$. The line passes through $P(1, 1, 3)$ with direction vector $\mathbf{d} = \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix}$.
Plane $\pi_1$ contains $A(0, -1, 3)$ and line $l$. The displacement vector $\vec{AP} = \begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix} - \begin{pmatrix} 0 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix}$.
The normal vector to $\pi_1$ is obtained using the cross product:
$\mathbf{n'} = \vec{AP} \times \mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \times \begin{pmatrix} -1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \\ 2 \end{pmatrix} = 2\begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$
We take the normal vector of the plane $\pi_1$ as $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$.

3. Calculate the acute angle between the line $AB$ and the plane $\pi_1$.

Let $\theta$ be the acute angle between the line $AB$ and the plane $\pi_1$. We use the relation $\sin \theta = \frac{|\mathbf{m} \cdot \mathbf{n}|}{|\mathbf{m}||\mathbf{n}|}$.
$\mathbf{m} \cdot \mathbf{n} = \begin{pmatrix} 6 \\ 3 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} = 6(2) + 3(-1) + (-2)(1) = 7$
$|\mathbf{m}| = \sqrt{6^2 + 3^2 + (-2)^2} = \sqrt{49} = 7$
$|\mathbf{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$
$\sin \theta = \frac{7}{(7)\sqrt{6}} = \frac{1}{\sqrt{6}}$
$\theta = \sin^{-1}\left(\frac{1}{\sqrt{6}}\right) \approx 24.0948^\circ$
Rounding to 1 decimal place, the acute angle is $24.1^\circ$.

1. Establish the general equation of a plane parallel to $\pi_2$.

Since the required planes are at a constant perpendicular distance from $\pi_2$, they must be parallel to $\pi_2$. Thus, they share the same normal vector $\mathbf{n} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$.
Let the cartesian equation of the required planes be $x + 2z = d$.

2. Set up the perpendicular distance expression from $\pi_2$ to the required planes.

By substituting $y = 0$ and $z = 0$ into the equation of $\pi_2$ ($x + 2z = 5$), we find that the point $(5, 0, 0)$ lies on $\pi_2$.
The perpendicular distance from the point $(5, 0, 0)$ to the plane $x + 2z - d = 0$ is:
$\text{Distance} = \frac{|1(5) + 0(0) + 2(0) - d|}{\sqrt{1^2 + 0^2 + 2^2}} = \frac{|5 - d|}{\sqrt{5}}$

3. Solve for the unknown constant $d$ using the given distance.

Equating the distance expression to the given distance of $2\sqrt{5}$:
$\frac{|5 - d|}{\sqrt{5}} = 2\sqrt{5}$
$|5 - d| = 2\sqrt{5} \times \sqrt{5}$
$|5 - d| = 10$
$5 - d = 10 \quad \text{or} \quad 5 - d = -10$
$d = -5 \quad \text{or} \quad d = 15$

4. State the cartesian equations of the required planes.

Substitute the values of $d$ back into the general equation.
The cartesian equations of the planes are $x + 2z = -5$ and $x + 2z = 15$.