直线 $l_1$ 和 $l_2$ 的方程分别为
$\frac{x-1}{3} = \frac{y-2}{a}, z = 1 \quad \text{与} \quad \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}, \lambda \in \mathbb{R}$
其中 $a$ 为常数。

1. Express the coordinates of a general point on $l_2$ using the parameter $\lambda$.

From the vector equation of $l_2$, a general point on $l_2$ has coordinates $(1 - \lambda, 2 + \lambda, \lambda)$.

2. Equate the $z$-coordinate of $l_2$ to the $z$-coordinate condition of $l_1$ to find $\lambda$.

Since $l_1$ and $l_2$ intersect at point $N$, the point $N$ must lie on $l_1$ and therefore satisfy its equation $z = 1$.
Substituting the $z$-coordinate of $l_2$ into this equation:
$\lambda = 1$

3. Find the coordinates of $N$.

Substituting $\lambda = 1$ back into the general coordinates of $l_2$:
$x = 1 - 1 = 0$
$y = 2 + 1 = 3$
$z = 1$
Therefore, the coordinates of $N$ are $(0, 3, 1)$.

4. Substitute the $x$ and $y$ coordinates of $N$ into the Cartesian equation of $l_1$ to find $a$.

Since $N(0, 3, 1)$ also lies on $l_1$, its coordinates must satisfy the given Cartesian equation $\frac{x-1}{3} = \frac{y-2}{a}$.
Substituting $x = 0$ and $y = 3$:
$\frac{0-1}{3} = \frac{3-2}{a}$
$-\frac{1}{3} = \frac{1}{a}$
$a = -3$

1. Define the position vector of F in terms of lambda and find the displacement vector PF.

Since $F$ lies on $l_2$, its position vector is $\vec{OF} = \begin{pmatrix} 1-\lambda \\ 2+\lambda \\ \lambda \end{pmatrix}$ for some $\lambda \in \mathbb{R}$.

The displacement vector $\vec{PF}$ is:
$\vec{PF} = \vec{OF} - \vec{OP} = \begin{pmatrix} 1-\lambda \\ 2+\lambda \\ \lambda \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -1-\lambda \\ 1+\lambda \\ \lambda-1 \end{pmatrix}$

2. Use the dot product to find the value of lambda, since PF is perpendicular to the line.

Since $\vec{PF}$ is perpendicular to $l_2$, the dot product of $\vec{PF}$ and the direction vector of $l_2$ is zero:
$\vec{PF} \cdot \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} = 0$
$\begin{pmatrix} -1-\lambda \\ 1+\lambda \\ \lambda-1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} = 0$
$(-1)(-1-\lambda) + (1)(1+\lambda) + (1)(\lambda-1) = 0$
$1 + \lambda + 1 + \lambda + \lambda - 1 = 0$
$3\lambda + 1 = 0 \implies \lambda = -\frac{1}{3}$

3. Substitute the value of lambda back into the position vector of F.

Substitute $\lambda = -\frac{1}{3}$ into $\vec{OF}$:
$\vec{OF} = \begin{pmatrix} 1 - (-\frac{1}{3}) \\ 2 + (-\frac{1}{3}) \\ -\frac{1}{3} \end{pmatrix} = \begin{pmatrix} \frac{4}{3} \\ \frac{5}{3} \\ -\frac{1}{3} \end{pmatrix}$
Hence, the position vector of $F$ is $\frac{4}{3}\mathbf{i} + \frac{5}{3}\mathbf{j} - \frac{1}{3}\mathbf{k}$ (shown).

1. Find the displacement vector $\vec{PF}$ using the coordinates of $F$ from part (ii)

$\vec{PF} = \vec{OF} - \vec{OP} = \begin{pmatrix} \frac{4}{3} \\ \frac{5}{3} \\ -\frac{1}{3} \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -\frac{2}{3} \\ \frac{2}{3} \\ -\frac{4}{3} \end{pmatrix}$

2. Calculate the magnitude of $\vec{PF}$ to find the exact distance

$\text{Distance} = |\vec{PF}| = \sqrt{\left(-\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(-\frac{4}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{4}{9} + \frac{16}{9}} = \sqrt{\frac{24}{9}} = \frac{\sqrt{24}}{3} = \frac{2\sqrt{6}}{3}$

1. Find a vector parallel to the plane by calculating the displacement vector from a point on the line to point $P$.

Let $A(1, 2, 0)$ be the point on $l_2$ derived from its position vector. The displacement vector from $A$ to $P(2, 1, 1)$ is:
$\vec{AP} = \vec{OP} - \vec{OA} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$

2. Calculate the normal vector to the plane using the cross product of the line's direction vector and $\vec{AP}$.

The plane $\pi$ contains $l_2$, so it is parallel to the direction vector of $l_2$, $\mathbf{d} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix}$.
A normal vector $\mathbf{n}$ to the plane is given by the cross product:
$\mathbf{n} = \mathbf{d} \times \vec{AP} = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} (1)(1) - (1)(-1) \\ (1)(1) - (-1)(1) \\ (-1)(-1) - (1)(1) \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}$
We can scale this to obtain a simpler normal vector, $\mathbf{n}' = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$.

3. Formulate the Cartesian equation of the plane.

Using the normal vector $\mathbf{n}' = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and the point $A(1, 2, 0)$ on the plane, the scalar product form is $\mathbf{r} \cdot \mathbf{n}' = \mathbf{a} \cdot \mathbf{n}'$:
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = (1)(1) + (2)(1) + (0)(0) = 3$
Converting this scalar product into a Cartesian equation by taking $\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$:
$x + y = 3$