平面 $p_1$ 和 $p_2$ 的方程分别由下式给出 $\mathbf{r} = s \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}, \quad s, t \in \mathbb{R} \quad \text{and} \quad \mathbf{r} \cdot \begin{pmatrix} \alpha \\ 3 \\ \beta \end{pmatrix} = 7,$ 其中 $\alpha$ 和 $\beta$ 为实常数。

1. Extract a point and the direction vector from the Cartesian equation of line $l_1$.

Rewrite the equation of $l_1$ into the standard form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$:

$l_1: \frac{x+1}{4} = \frac{y-2}{-3} = \frac{z-0}{1}$

From this, a point on $l_1$ has position vector $\mathbf{a} = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}$ and the direction vector of $l_1$ is $\mathbf{d} = \begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix}$.

2. Substitute the point on $l_1$ into the equation of plane $p_2$ to find $\alpha$.

Since the line $l_1$ lies completely in plane $p_2$, any point on $l_1$ must also lie on $p_2$. Substituting the point $(-1, 2, 0)$ into the equation of $p_2$:

$\begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} \alpha \\ 3 \\ \beta \end{pmatrix} = 7$

$-\alpha + 6 + 0 = 7$
$-\alpha = 1$
$\alpha = -1$

3. Use the perpendicularity between the line's direction vector and the plane's normal vector to find $\beta$.

Since $l_1$ lies in $p_2$, its direction vector $\mathbf{d}$ must be perpendicular to the normal vector $\mathbf{n_2}$ of $p_2$. Hence, their dot product is zero:

$\mathbf{d} \cdot \mathbf{n_2} = 0$
$\begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 3 \\ \beta \end{pmatrix} = 0$

$-4 - 9 + \beta = 0$
$-13 + \beta = 0$
$\beta = 13$

1. Substitute the parametric form of $p_1$ into the scalar product equation of $p_2$ to find the relationship between parameters $s$ and $t$.

From the equation of $p_1$, a generic point on the plane has position vector:
$\mathbf{r} = \begin{pmatrix} 3t \\ s - t \\ s + 2t \end{pmatrix}$
Substituting this into the equation of $p_2$:
$\begin{pmatrix} 3t \\ s - t \\ s + 2t \end{pmatrix} \cdot \begin{pmatrix} -3 \\ 3 \\ 2 \end{pmatrix} = 7$
$-3(3t) + 3(s - t) + 2(s + 2t) = 7$
$-9t + 3s - 3t + 2s + 4t = 7$
$5s - 8t = 7$

2. Express $s$ in terms of $t$ and substitute it back into the equation of $p_1$.

$s = \frac{8t + 7}{5}$
Substituting $s$ back into the equation of $p_1$:
$\mathbf{r} = \left( \frac{8t + 7}{5} \right) \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$
$\mathbf{r} = \begin{pmatrix} 0 \\ \frac{7}{5} \\ \frac{7}{5} \end{pmatrix} + \frac{8t}{5} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix}$
$\mathbf{r} = \begin{pmatrix} 0 \\ \frac{7}{5} \\ \frac{7}{5} \end{pmatrix} + \frac{t}{5} \begin{pmatrix} 15 \\ 8 - 5 \\ 8 + 10 \end{pmatrix}$
$\mathbf{r} = \begin{pmatrix} 0 \\ \frac{7}{5} \\ \frac{7}{5} \end{pmatrix} + \frac{t}{5} \begin{pmatrix} 15 \\ 3 \\ 18 \end{pmatrix}$

3. Simplify the direction vector to obtain the standard vector equation of the line.

Factoring out $3$ from the direction vector components:
$\mathbf{r} = \begin{pmatrix} 0 \\ \frac{7}{5} \\ \frac{7}{5} \end{pmatrix} + \frac{3t}{5} \begin{pmatrix} 5 \\ 1 \\ 6 \end{pmatrix}$
Letting $\lambda = \frac{3t}{5}$, the line of intersection is:
$\mathbf{r} = \begin{pmatrix} 0 \\ \frac{7}{5} \\ \frac{7}{5} \end{pmatrix} + \lambda \begin{pmatrix} 5 \\ 1 \\ 6 \end{pmatrix}, \quad \lambda \in \mathbb{R}$

1. Identify the coordinates of a known point on the line $l_1$.

From the equation of $l_1: \frac{x+1}{4} = \frac{y-2}{-3} = \frac{z}{1}$, a point on the line is $A(-1, 2, 0)$.

2. Calculate the perpendicular distance from point $A$ to plane $p_3$.

Since $l_1$ is parallel to $p_3$, the distance between them is the perpendicular distance from any point on $l_1$ to $p_3$.

The equation of $p_3$ can be expressed as $\mathbf{r} \cdot \begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix} = -4$.
Using the distance formula $\frac{|\mathbf{a} \cdot \mathbf{n} - d|}{|\mathbf{n}|}$:

$\text{Distance} = \frac{\left| \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 1 \\ 3 \end{pmatrix} - (-4) \right|}{\sqrt{0^2 + 1^2 + 3^2}}$
$\text{Distance} = \frac{|(0 + 2 + 0) + 4|}{\sqrt{10}}$
$\text{Distance} = \frac{6}{\sqrt{10}} = \frac{3\sqrt{10}}{5}$