分别由方程 $2x+y=1$ 和 $8x+ay+z=4$ 给出的两个平面 $p_{1}$ 和 $p_{2}$ 相交于一条直线 $L$，该直线包含点 $A(0,1,0)$。

1. Substitute the point $A$ into the equation of plane $p_2$ to show $a=4$.

Since the line $L$ contains $A(0,1,0)$, the point $A$ must lie on both planes $p_1$ and $p_2$.
Substituting $x=0$, $y=1$, $z=0$ into the equation of $p_2$ ($8x+ay+z=4$):
$8(0) + a(1) + 0 = 4$
$a = 4 \text{ (shown)}$

2. Find the direction vector of line $L$.

The line of intersection $L$ is perpendicular to the normal vectors of both planes.
The normal vectors are $\mathbf{n}_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$ for $p_1$ and $\mathbf{n}_2 = \begin{pmatrix} 8 \\ 4 \\ 1 \end{pmatrix}$ for $p_2$.
The direction vector $\mathbf{d}$ of line $L$ is given by the cross product:
$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 8 \\ 4 \\ 1 \end{pmatrix} = \begin{pmatrix} (1)(1) - (0)(4) \\ (0)(8) - (2)(1) \\ (2)(4) - (1)(8) \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$

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

Using the point $A(0,1,0)$ on the line and the direction vector $\mathbf{d}$:
$\mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}, \lambda \in \mathbb{R}$

1. Set up equations using the coordinates of $B$ to determine the direction of $\overrightarrow{AB}$.

Let the position vector of $B$ be $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$.
Since $B$ lies in plane $p_1$ with equation $2x + y = 1$, we can express $y$ as $y = 1 - 2x$.
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} x \\ y - 1 \\ z \end{pmatrix} = \begin{pmatrix} x \\ -2x \\ z \end{pmatrix}$
From part (i), the direction vector of line $L$ is $\mathbf{d} = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$.
Since $AB$ is perpendicular to $L$, their dot product must be zero:
$\overrightarrow{AB} \cdot \mathbf{d} = \begin{pmatrix} x \\ -2x \\ z \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} = 0$
$x(1) + (-2x)(-2) + z(0) = 0 \implies 5x = 0 \implies x = 0$
Substituting $x = 0$ back into $\overrightarrow{AB}$ yields:
$\overrightarrow{AB} = \begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix} = z \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$
Thus, $\overrightarrow{AB}$ is parallel to $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

2. Express the position vector of $B$ in terms of a single parameter.

Since $x = 0$, we have $y = 1 - 2(0) = 1$. Let the $z$-coordinate of $B$ be a parameter $k$.
Therefore, the position vector of $B$ is $\overrightarrow{OB} = \begin{pmatrix} 0 \\ 1 \\ k \end{pmatrix}$.

3. Apply the perpendicular distance formula from point $B$ to plane $p_2$.

The equation of plane $p_2$ from part (i) is $8x + 4y + z = 4$.
The perpendicular distance from $B(0, 1, k)$ to $p_2$ is given as $5$ units.
Using the point-to-plane distance formula:
$\text{Distance} = \frac{|8(0) + 4(1) + (k) - 4|}{\sqrt{8^2 + 4^2 + 1^2}} = 5$
$\frac{|4 + k - 4|}{\sqrt{64 + 16 + 1}} = 5$
$\frac{|k|}{\sqrt{81}} = 5 \implies \frac{|k|}{9} = 5$

4. Solve for the parameter to find the possible position vectors of $B$.

$|k| = 45 \implies k = 45 \text{ or } k = -45$
Therefore, the possible position vectors of $B$ are $\begin{pmatrix} 0 \\ 1 \\ 45 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 1 \\ -45 \end{pmatrix}$.

1. Identify the direction vector of the line and the normal vector of the plane from earlier parts.

From part (i), the equation of plane $p_2$ is $8x + 4y + z = 4$, so its normal vector is $\mathbf{n_2} = \begin{pmatrix} 8 \\ 4 \\ 1 \end{pmatrix}$.

From part (ii), the direction vector of line $AB$ is parallel to $\mathbf{d} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.

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

Let $\theta$ be the acute angle between line $AB$ and plane $p_2$.

$\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n_2}|}{|\mathbf{d}| |\mathbf{n_2}|}$
$\sin \theta = \frac{\left| \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 8 \\ 4 \\ 1 \end{pmatrix} \right|}{\sqrt{0^2 + 0^2 + 1^2} \sqrt{8^2 + 4^2 + 1^2}}$
$\sin \theta = \frac{1}{(1)(\sqrt{81})} = \frac{1}{9}$
$\theta = \sin^{-1}\left(\frac{1}{9}\right) \approx 6.379^\circ$

The acute angle is $6.4^\circ$ (to 1 d.p.).

1. Set up a system of equations to test for the existence of a common intersection point.

To find any common point, we attempt to solve the system of Cartesian equations for the three planes. Using $a=4$ from part (i):
$\begin{cases} 2x+y=1 & \text{--- (1)} \\ 8x+4y+z=4 & \text{--- (2)} \\ 2x+y+\beta z=6 & \text{--- (3)} \end{cases}$

2. Attempt to solve the system and show that a contradiction arises for any $\beta$.

Substitute $(1)$ into $(2)$:
$4(2x+y) + z = 4 \implies 4(1) + z = 4 \implies z = 0$

Substitute $(1)$ and $z = 0$ into $(3)$:
$1 + \beta(0) = 6$
$1 = 6$

The result $1 = 6$ is a mathematical contradiction that holds regardless of the value of $\beta$. Therefore, the system has no solution, meaning the three planes do not meet at a common point for all values of $\beta$.

3. Determine the geometrical relationship of the planes when $\beta=0$.

When $\beta=0$, the equation of $p_3$ simplifies to $2x+y=6$.
The equation of $p_1$ is $2x+y=1$.

Since both planes have the same normal vector $\mathbf{n} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$ but different scalar constants ($1 \neq 6$), $p_1$ and $p_3$ are parallel and distinct planes.
(Because $p_2$ is not parallel to them, the three planes geometrically consist of two parallel planes intersected by a third plane).