平面 $p_1$ 和 $p_2$ 的方程由下式给出： $p_1:\;\mathbf{r}\cdot \begin{pmatrix}\alpha\\7\\2\end{pmatrix} = \alpha + 20,\; \alpha \in \mathbb{R},$ $p_2:\;\mathbf{r}\cdot \begin{pmatrix}3\\5\\\beta\end{pmatrix} = 7,\; \beta \in \mathbb{R}.$

1. Verify that point $A$ lies on plane $p_1$ by substituting its position vector into the equation of $p_1$.

Substitute the position vector of $A$, $\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix}$, into the LHS of the equation for $p_1$:

$\text{LHS} = \begin{pmatrix}1\\2\\3\end{pmatrix} \cdot \begin{pmatrix}\alpha\\7\\2\end{pmatrix}$

$\text{LHS} = 1(\alpha) + 2(7) + 3(2)$

$\text{LHS} = \alpha + 14 + 6 = \alpha + 20$

Since the LHS evaluates to $\alpha + 20$, which matches the RHS of the equation of $p_1$, the point $A(1,2,3)$ lies in $p_1$.

2. Find the value of $\beta$ by substituting the position vector of $A$ into the equation of $p_2$.

Since $A$ also lies in $p_2$, its position vector must satisfy the equation of $p_2$:

$\begin{pmatrix}1\\2\\3\end{pmatrix} \cdot \begin{pmatrix}3\\5\\\beta\end{pmatrix} = 7$

$1(3) + 2(5) + 3(\beta) = 7$

$3 + 10 + 3\beta = 7$

$13 + 3\beta = 7$

$3\beta = -6$

$\beta = -2$

1. Determine the normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$ using information from earlier parts.

From part (ii), we use the shown value $\alpha = 3$, giving the normal vector of $p_1$ as $\mathbf{n}_1 = \begin{pmatrix} 3 \\ 7 \\ 2 \end{pmatrix}$.

To find $\beta$, we substitute the coordinates of point $A(1,2,3)$ into the equation of $p_2$:
$\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 5 \\ \beta \end{pmatrix} = 7$
$3(1) + 5(2) + 3\beta = 7$
$13 + 3\beta = 7 \implies \beta = -2$

Thus, the normal vector of $p_2$ is $\mathbf{n}_2 = \begin{pmatrix} 3 \\ 5 \\ -2 \end{pmatrix}$.

2. Calculate the acute angle between the planes using the scalar product of their normal vectors.

The acute angle $\theta$ between the two planes is given by:
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$
$\cos \theta = \frac{\left| \begin{pmatrix} 3 \\ 7 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 5 \\ -2 \end{pmatrix} \right|}{\sqrt{3^2 + 7^2 + 2^2} \sqrt{3^2 + 5^2 + (-2)^2}}$
$\cos \theta = \frac{|3(3) + 7(5) + 2(-2)|}{\sqrt{62}\sqrt{38}}$
$\cos \theta = \frac{40}{\sqrt{2356}}$
$\theta = \cos^{-1}\left(\frac{40}{\sqrt{2356}}\right) \approx 34.505^\circ$

Rounding to 1 decimal place, $\theta = 34.5^\circ$.

1. Formulate the vector from $B$ to a general point $F$ on line $l$.

From part (ii), the equation of line $l$ is $\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + k\begin{pmatrix}1\\-2\\1\end{pmatrix}, k \in \mathbb{R}$.
Let $F$ be the foot of the perpendicular from $B(2, -4, 8)$ to $l$. For some parameter $k$, the position vector is:
$\vec{OF} = \begin{pmatrix}1+k \\ 2-2k \\ 3+k\end{pmatrix}$
The vector from $B$ to $F$ is:
$\vec{BF} = \vec{OF} - \vec{OB} = \begin{pmatrix}1+k \\ 2-2k \\ 3+k\end{pmatrix} - \begin{pmatrix}2 \\ -4 \\ 8\end{pmatrix} = \begin{pmatrix}k-1 \\ 6-2k \\ k-5\end{pmatrix}$

2. Apply the perpendicularity condition to find the value of $k$.

Since $\vec{BF}$ is perpendicular to the direction vector of $l$, their scalar product must be zero:
$\vec{BF} \cdot \mathbf{d} = 0$
$\begin{pmatrix}k-1 \\ 6-2k \\ k-5\end{pmatrix} \cdot \begin{pmatrix}1 \\ -2 \\ 1\end{pmatrix} = 0$
$(k-1)(1) + (6-2k)(-2) + (k-5)(1) = 0$
$k - 1 - 12 + 4k + k - 5 = 0$
$6k - 18 = 0 \implies k = 3$

3. Calculate the position vector of the foot of the perpendicular.

Substitute $k = 3$ back into $\vec{OF}$:
$\vec{OF} = \begin{pmatrix}1+3 \\ 2-2(3) \\ 3+3\end{pmatrix} = \begin{pmatrix}4 \\ -4 \\ 6\end{pmatrix}$

4. Find the position vector of the reflection of $B$ about line $l$.

Let $B'$ be the reflection of $B$ about line $l$. The foot of the perpendicular $F$ is the midpoint of $BB'$:
$\vec{OF} = \frac{\vec{OB} + \vec{OB'}}{2} \implies \vec{OB'} = 2\vec{OF} - \vec{OB}$
$\vec{OB'} = 2\begin{pmatrix}4 \\ -4 \\ 6\end{pmatrix} - \begin{pmatrix}2 \\ -4 \\ 8\end{pmatrix} = \begin{pmatrix}8 \\ -8 \\ 12\end{pmatrix} - \begin{pmatrix}2 \\ -4 \\ 8\end{pmatrix} = \begin{pmatrix}6 \\ -4 \\ 4\end{pmatrix}$

5. Find the equation of the reflected line.

The original line passes through $A$ and $B$. From part (i), $A(1, 2, 3)$ lies on both planes and thus lies on their intersection, line $l$. Because $A$ is on the axis of reflection, its reflection is itself.
Therefore, the reflected line passes through $A$ and $B'$. The direction vector of the reflected line is:
$\vec{AB'} = \vec{OB'} - \vec{OA} = \begin{pmatrix}6 \\ -4 \\ 4\end{pmatrix} - \begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} = \begin{pmatrix}5 \\ -6 \\ 1\end{pmatrix}$
The vector equation of the reflected line is:
$\mathbf{r} = \begin{pmatrix}1 \\ 2 \\ 3\end{pmatrix} + \lambda\begin{pmatrix}5 \\ -6 \\ 1\end{pmatrix}, \lambda \in \mathbb{R}$