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 in $p_1$ by substituting its position vector into the equation of $p_1$.

Let the position vector of $A$ be $\mathbf{r}_A = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$.
Substituting $\mathbf{r}_A$ into the LHS of the equation for $p_1$:
$\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} \alpha \\ 7 \\ 2 \end{pmatrix} = 1(\alpha) + 2(7) + 3(2)$
$= \alpha + 14 + 6 = \alpha + 20$
Since the result matches the RHS ($\alpha + 20$), the point $A$ satisfies the equation of $p_1$ and thus lies in $p_1$.

2. Substitute the position vector of $A$ into the equation of $p_2$ to find $\beta$.

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. Retrieve the values of $\alpha$ and $\beta$ from earlier parts to determine the normal vectors of $p_1$ and $p_2$.

From part (ii), it is given that $\alpha = 3$.
For $\beta$, using the information from part (i) that point $A(1, 2, 3)$ lies in $p_2$:
$\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 5 \\ \beta \end{pmatrix} = 7$
$3(1) + 5(2) + \beta(3) = 7$
$13 + 3\beta = 7 \implies 3\beta = -6 \implies \beta = -2$
Thus, the normal vectors to the planes $p_1$ and $p_2$ are:
$\mathbf{n}_1 = \begin{pmatrix} 3 \\ 7 \\ 2 \end{pmatrix}, \quad \mathbf{n}_2 = \begin{pmatrix} 3 \\ 5 \\ -2 \end{pmatrix}$

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

Let $\theta$ be the acute angle between $p_1$ and $p_2$.
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$
Calculating the scalar product and the magnitudes:
$\mathbf{n}_1 \cdot \mathbf{n}_2 = \begin{pmatrix} 3 \\ 7 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 5 \\ -2 \end{pmatrix} = 3(3) + 7(5) + 2(-2) = 9 + 35 - 4 = 40$
$|\mathbf{n}_1| = \sqrt{3^2 + 7^2 + 2^2} = \sqrt{9 + 49 + 4} = \sqrt{62}$
$|\mathbf{n}_2| = \sqrt{3^2 + 5^2 + (-2)^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$
Substituting these values into the formula:
$\cos \theta = \frac{40}{\sqrt{62}\sqrt{38}} = \frac{40}{\sqrt{2356}}$
$\theta = \cos^{-1}\left(\frac{40}{\sqrt{2356}}\right) \approx 34.522^\circ$
Therefore, the acute angle is $34.5^\circ$ (to 1 decimal place).

1. Express the position vector of the foot of the perpendicular and the vector $\vec{BF}$ in terms of $k$.

From part (ii), the equation of line $l$ is $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + k \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$.
Let the foot of the perpendicular from $B$ to $l$ be $F$. Since $F$ lies on $l$, its position vector is $\vec{OF} = \begin{pmatrix} 1+k \\ 2-2k \\ 3+k \end{pmatrix}$ for some value of $k$.
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. Use the perpendicular condition to solve for $k$ and find the position vector of $F$.

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

Substitute $k = 3$ to find the position vector of the foot of the perpendicular:
$\vec{OF} = \begin{pmatrix} 1+3 \\ 2-6 \\ 3+3 \end{pmatrix} = \begin{pmatrix} 4 \\ -4 \\ 6 \end{pmatrix}$

3. Find the position vector of $B'$, the reflection of $B$ about $l$.

The point $A(1, 2, 3)$ lies on $l$, so its reflection about $l$ is itself.
Let the reflection of point $B$ about $l$ be $B'$. The foot of the perpendicular $F$ is the midpoint of the line segment $BB'$.
$\vec{OF} = \frac{\vec{OB} + \vec{OB'}}{2}$
$\vec{OB'} = 2\vec{OF} - \vec{OB}$
$\vec{OB'} = 2\begin{pmatrix} 4 \\ -4 \\ 6 \end{pmatrix} - \begin{pmatrix} 2 \\ -4 \\ 8 \end{pmatrix} = \begin{pmatrix} 8 \\ -8 \\ 12 \end{pmatrix} - \begin{pmatrix} 2 \\ -4 \\ 8 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \\ 4 \end{pmatrix}$

4. Form the equation of the reflected line.

The reflected line passes through $A$ and $B'$. Its direction vector 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 equation of the reflected line is:
$\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ -6 \\ 1 \end{pmatrix}, \; \lambda \in \mathbb{R}$