两条直线 $l_1$ 和 $l_2$ 的方程分别为 $\frac{x - 1}{2} = -z - 1, y = 2$ 与 $x + 3 = 4 - y = z + 2$。

1. Express the equations of $l_1$ and $l_2$ in parametric form.

Let $l_1$ be defined by $\frac{x-1}{2} = -(z+1) = \lambda$ and $y=2$.
$x = 1 + 2\lambda, \quad y = 2, \quad z = -1 - \lambda$

Let $l_2$ be defined by $x+3 = 4-y = z+2 = \mu$.
$x = -3 + \mu, \quad y = 4 - \mu, \quad z = -2 + \mu$

2. Set up a system of equations by equating the $x$, $y$, and $z$ coordinates.

For the lines to intersect, they must share a common point for some values of $\lambda$ and $\mu$:
$1 + 2\lambda = -3 + \mu \quad \text{--- (1)}$
$2 = 4 - \mu \quad \text{--- (2)}$
$-1 - \lambda = -2 + \mu \quad \text{--- (3)}$

3. Solve for the parameters and verify consistency across all three equations.

From (2), we find:
$\mu = 2$

Substitute $\mu = 2$ into (1):
$1 + 2\lambda = -3 + 2$
$2\lambda = -2 \implies \lambda = -1$

Check consistency by substituting $\lambda = -1$ and $\mu = 2$ into (3):
$\text{LHS} = -1 - (-1) = 0$
$\text{RHS} = -2 + 2 = 0$
Since LHS = RHS, the system of equations is consistent, which means $l_1$ and $l_2$ intersect.

4. Calculate the coordinates of the intersection point.

Substitute $\mu = 2$ into the parametric equations for $l_2$:
$x = -3 + 2 = -1$
$y = 4 - 2 = 2$
$z = -2 + 2 = 0$

(Alternatively, substituting $\lambda = -1$ into $l_1$ yields the same result).
The intersection point is $(-1, 2, 0)$.

1. Extract the direction vectors of lines $l_1$ and $l_2$ from their Cartesian equations.

For $l_1$, the Cartesian equation is $\frac{x - 1}{2} = \frac{z + 1}{-1}, y = 2$.
The direction vector is $\mathbf{d}_1 = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}$.

For $l_2$, the equation is $x + 3 = \frac{y - 4}{-1} = z + 2$.
The direction vector is $\mathbf{d}_2 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$.

2. Use the dot product to find the acute angle between the two lines.

Let $\theta$ be the acute angle between $l_1$ and $l_2$.
$\cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1||\mathbf{d}_2|}$
$\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (0)(-1) + (-1)(1) = 1$
$|\mathbf{d}_1| = \sqrt{2^2 + 0^2 + (-1)^2} = \sqrt{5}$
$|\mathbf{d}_2| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$
$\cos \theta = \frac{|1|}{\sqrt{5}\sqrt{3}} = \frac{1}{\sqrt{15}}$
$\theta = \cos^{-1}\left(\frac{1}{\sqrt{15}}\right) \approx 75.036^\circ$

1. Find the point of intersection $A$ of lines $l_1$ and $l_2$.

Express $l_1$ and $l_2$ in vector form:
$l_1: \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}$
$l_2: \mathbf{r} = \begin{pmatrix} -3 \\ 4 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$

Equating coordinates to find the intersection:
$1 + 2\lambda = -3 + \mu \quad \text{--- (1)}$
$2 = 4 - \mu \implies \mu = 2 \quad \text{--- (2)}$

Substitute $\mu = 2$ into (1): $1 + 2\lambda = -1 \implies \lambda = -1$.
Substitute $\lambda = -1$ into the equation of $l_1$ to find the intersection point $A$:
$\vec{OA} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}$
The point of intersection is $A(-1, 2, 0)$.

2. Define a point $B$ on $l_1$ and find the displacement vector $\vec{BF}$ to a generic point $F$ on $l_2$.

Let $B(1, 2, -1)$ be the position on $l_1$ when $\lambda = 0$.
Let $F$ be the foot of the perpendicular from $B$ to $l_2$. The position vector of $F$ is:
$\vec{OF} = \begin{pmatrix} -3 + \mu \\ 4 - \mu \\ -2 + \mu \end{pmatrix}$

The displacement vector $\vec{BF}$ is:
$\vec{BF} = \vec{OF} - \vec{OB} = \begin{pmatrix} -3 + \mu \\ 4 - \mu \\ -2 + \mu \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} \mu - 4 \\ 2 - \mu \\ \mu - 1 \end{pmatrix}$

3. Use the scalar product to find the foot of the perpendicular, $F$, from $B$ to $l_2$.

Since $\vec{BF}$ is perpendicular to $l_2$, its scalar product with the direction vector of $l_2$ must be zero:
$\begin{pmatrix} \mu - 4 \\ 2 - \mu \\ \mu - 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = 0$
$(\mu - 4) - (2 - \mu) + (\mu - 1) = 0$
$3\mu - 7 = 0 \implies \mu = \frac{7}{3}$

Substitute $\mu = \frac{7}{3}$ to find $\vec{OF}$:
$\vec{OF} = \begin{pmatrix} -3 + 7/3 \\ 4 - 7/3 \\ -2 + 7/3 \end{pmatrix} = \begin{pmatrix} -2/3 \\ 5/3 \\ 1/3 \end{pmatrix}$

4. Find the position vector of the reflected point $B'$ using the midpoint.

The point $F$ is the midpoint of $B$ and its reflection $B'$ across $l_2$:
$\vec{OF} = \frac{\vec{OB} + \vec{OB'}}{2} \implies \vec{OB'} = 2\vec{OF} - \vec{OB}$

$\vec{OB'} = 2 \begin{pmatrix} -2/3 \\ 5/3 \\ 1/3 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} -4/3 \\ 10/3 \\ 2/3 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} -7/3 \\ 4/3 \\ 5/3 \end{pmatrix}$

5. Determine the direction vector of $l_3$ and form its equation.

The reflected line $l_3$ passes through the common intersection point $A$ and the reflected point $B'$.
The direction vector of $l_3$ is $\vec{AB'} = \vec{OB'} - \vec{OA}$:
$\vec{AB'} = \begin{pmatrix} -7/3 \\ 4/3 \\ 5/3 \end{pmatrix} - \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} -4/3 \\ -2/3 \\ 5/3 \end{pmatrix}$

To avoid fractions, we can multiply the vector by $-3$ to get a parallel direction vector for $l_3$:
$\mathbf{d}_3 = \begin{pmatrix} 4 \\ 2 \\ -5 \end{pmatrix}$

The vector equation of $l_3$ is $\mathbf{r} = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} + t \begin{pmatrix} 4 \\ 2 \\ -5 \end{pmatrix}$, where $t \in \mathbb{R}$.
Converting this to Cartesian form yields:
$\frac{x + 1}{4} = \frac{y - 2}{2} = \frac{z}{-5}$