以原点 $O$ 为基准，点 $A$、$B$ 和 $C$ 的位置向量分别为 $3\mathbf{i} - \mathbf{j}$、$5\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$ 和 $2\mathbf{i} + \alpha\mathbf{j} + \sqrt{2}\mathbf{k}$，其中 $\alpha$ 是一个非零常数。

1. Find the direction vector of the line passing through $A$ and $C$.

$\vec{AC} = \vec{OC} - \vec{OA} = \begin{pmatrix} 2 \\ \alpha \\ \sqrt{2} \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ \alpha + 1 \\ \sqrt{2} \end{pmatrix}$

2. Set up an equation using the acute angle between $\vec{AC}$ and the $x$-axis.

The direction vector of the $x$-axis is $\mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$.
The acute angle $\theta$ between a line and the $x$-axis is given by $\cos \theta = \frac{|\mathbf{d} \cdot \mathbf{i}|}{|\mathbf{d}| |\mathbf{i}|}$:
$\cos 60^\circ = \frac{\left| \begin{pmatrix} -1 \\ \alpha + 1 \\ \sqrt{2} \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right|}{\sqrt{(-1)^2 + (\alpha + 1)^2 + (\sqrt{2})^2} (1)}$
$\frac{1}{2} = \frac{|-1|}{\sqrt{1 + (\alpha + 1)^2 + 2}}$
$\frac{1}{2} = \frac{1}{\sqrt{3 + (\alpha + 1)^2}}$

3. Solve the equation to find the value of $\alpha$.

$\sqrt{3 + (\alpha + 1)^2} = 2$
Squaring both sides:
$3 + (\alpha + 1)^2 = 4$
$(\alpha + 1)^2 = 1$
$\alpha + 1 = 1 \quad \text{or} \quad \alpha + 1 = -1$
$\alpha = 0 \quad \text{or} \quad \alpha = -2$
Since the question states that $\alpha$ is a non-zero constant, we reject $\alpha = 0$.
$\alpha = -2$

1. Find the displacement vector $\vec{AB}$ to determine the direction of the line.

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 5 \\ 3 \\ 4 \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 4 \\ 4 \end{pmatrix}$

2. Calculate the magnitude of $\vec{AB}$ and deduce its unit vector.

$|\vec{AB}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{36} = 6$
Unit vector in the direction of $\vec{AB} = \frac{1}{6} \begin{pmatrix} 2 \ 4 \ 4 \end{pmatrix} = \begin{pmatrix} 1/3 \ 2/3 \ 2/3 \end{pmatrix} $$

3. Express the displacement vector $\vec{AQ}$ using the given distance.

Since $Q$ lies on $l$ and $AQ = 3$, the displacement $\vec{AQ}$ is $\pm 3$ times the unit vector in the direction of $\vec{AB}$.
$\vec{AQ} = \pm 3 \begin{pmatrix} 1/3 \\ 2/3 \\ 2/3 \end{pmatrix} = \pm \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$

4. Calculate the two possible position vectors for $Q$.

Using $\vec{OQ} = \vec{OA} + \vec{AQ}$:
$\vec{OQ}_1 = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix} + \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 2 \end{pmatrix}$
$\vec{OQ}_2 = \begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -3 \\ -2 \end{pmatrix}$

1. Form the equation of the line passing through $B$ and perpendicular to the plane $\pi$.

The plane $\pi$ has equation $x + 2z = 3$, so its normal vector is $\mathbf{n} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$.

The line perpendicular to $\pi$ passing through $B(5, 3, 4)$ has equation:
$\mathbf{r} = \begin{pmatrix} 5 \\ 3 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 + \lambda \\ 3 \\ 4 + 2\lambda \end{pmatrix}$

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

Since the point $F$ lies on $\pi$, its coordinates must satisfy the plane equation $x + 2z = 3$:
$(5 + \lambda) + 2(4 + 2\lambda) = 3$
$5 + \lambda + 8 + 4\lambda = 3$
$13 + 5\lambda = 3$
$5\lambda = -10 \implies \lambda = -2$

3. Calculate the position vector of $F$.

Substitute $\lambda = -2$ back into the line equation to find $\vec{OF}$:
$\vec{OF} = \begin{pmatrix} 5 - 2 \\ 3 \\ 4 + 2(-2) \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 0 \end{pmatrix}$
Hence, the position vector of $F$ is $3\mathbf{i} + 3\mathbf{j}$.

1. Calculate the position vector of F, the point on $\pi$ closest to B (required from part c).

\text{The line through } B \text{ perpendicular to } \pi \text{ is } \mathbf{r} = \begin{pmatrix} 5 \ 3 \ 4 \end{pmatrix} + \lambda \begin{pmatrix} 1 \ 0 \ 2 \end{pmatrix}. \
\text{Substituting into the equation of } \pi \ (x + 2z = 3): \
(5+\lambda) + 2(4+2\lambda) = 3 \implies 13 + 5\lambda = 3 \implies \lambda = -2. \
\vec{OF} = \begin{pmatrix} 5-2 \ 3 \ 4-4 \end{pmatrix} = \begin{pmatrix} 3 \ 3 \ 0 \end{pmatrix}.

2. Find the position vector of $B'$, the reflection of $B$ in $\pi$.

\text{Let } B' \text{ be the reflection of } B \text{ in } \pi. \text{ Since } F \text{ is the midpoint of } B \text{ and } B': \
\vec{OF} = \frac{\vec{OB} + \vec{OB'}}{2} \
\vec{OB'} = 2\vec{OF} - \vec{OB} = 2\begin{pmatrix} 3 \ 3 \ 0 \end{pmatrix} - \begin{pmatrix} 5 \ 3 \ 4 \end{pmatrix} = \begin{pmatrix} 1 \ 3 \ -4 \end{pmatrix}.

3. Identify the point of intersection between line $l$ and plane $\pi$.

\text{Line } l \text{ passes through } A \text{ and } B. \
\text{The plane } \pi \text{ contains } A(3, -1, 0), \text{ so } A \text{ is the point of intersection between } l \text{ and } \pi. \
\text{Therefore, the reflected line } l' \text{ passes through } A \text{ and } B'.

4. Form the vector equation of the reflected line.

\text{The direction vector of } l' \text{ is } \vec{AB'} = \begin{pmatrix} 1 \ 3 \ -4 \end{pmatrix} - \begin{pmatrix} 3 \ -1 \ 0 \end{pmatrix} = \begin{pmatrix} -2 \ 4 \ -4 \end{pmatrix}. \
\text{Using a parallel direction vector } \mathbf{d} = \begin{pmatrix} -1 \ 2 \ -2 \end{pmatrix}, \
\text{The vector equation of } l' \text{ is } \mathbf{r} = \begin{pmatrix} 3 \ -1 \ 0 \end{pmatrix} + \mu \begin{pmatrix} -1 \ 2 \ -2 \end{pmatrix}, \mu \in \mathbb{R}.