平面 $p_1$ 的方程为 $x - y + z = 0$，平面 $p_2$ 的方程为 $x + y + z = 2$。 以原点 $O$ 为基准，点 $A$ 和点 $B$ 满足 $\overrightarrow{OA} = -2\mathbf{i} + \mathbf{j} - \mathbf{k}$ 且 $\overrightarrow{OB} = \mathbf{i} + \mathbf{j}$。直线 $l$ 的方程为 $\mathbf{r} = -2\mathbf{i} + \mathbf{j} - \mathbf{k} + \mu(\mathbf{i} - \mathbf{k})$，其中 $\mu$ 为参数。

1. Identify the normal vectors of the planes from their cartesian equations.

For plane $p_1: x - y + z = 0$, the normal vector is $\mathbf{n}_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$.

For plane $p_2: x + y + z = 2$, the normal vector is $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.

2. Calculate the scalar product and the magnitudes of the normal vectors.

$\mathbf{n}_1 \cdot \mathbf{n}_2 = (1)(1) + (-1)(1) + (1)(1) = 1 - 1 + 1 = 1$

$|\mathbf{n}_1| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$
$|\mathbf{n}_2| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

3. Determine the acute angle between the two planes.

Let $\theta$ be the acute angle between the planes. The angle between the planes is given by the angle between their normal vectors:

$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$
$\cos \theta = \frac{|1|}{\sqrt{3}\sqrt{3}} = \frac{1}{3}$

$\theta = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.528...^\circ$

$\theta = 70.5^\circ$ (to 1 d.p.)

1. Solve the simultaneous Cartesian equations of the two planes to establish a relationship between the coordinates.

$x - y + z = 0 \quad \text{--- (1)}$
$x + y + z = 2 \quad \text{--- (2)}$
Adding (1) and (2) eliminates $y$:
$2x + 2z = 2 \implies x = 1 - z$
Subtracting (1) from (2) eliminates $x$ and $z$:
$2y = 2 \implies y = 1$

2. Introduce a parameter to express the line of intersection in vector form.

Let $z = \lambda$, where $\lambda \in \mathbb{R}$.
Substitute $z = \lambda$ into the relationships found:
$x = 1 - \lambda$
$y = 1$

Hence, the position vector $\mathbf{r}$ of any point on the line is:
$\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 - \lambda \\ 1 \\ \lambda \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$

1. Convert the given cartesian equation of the new line into vector form to identify a point on it and its direction vector.

The cartesian equation is $x - 3 = -z + 2, \; y = 1$.
This can be rewritten as $\frac{x - 3}{1} = \frac{z - 2}{-1}, \; y = 1$.
Let this be line $m$. A point on $m$ is $P(3, 1, 2)$ and its direction vector is $\mathbf{d} = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$.

2. Identify the relationship between line $l$ and line $m$, and find a vector connecting a point on $l$ to $P$.

Line $l$ has equation $\mathbf{r} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$.
Since both lines share the same direction vector $\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$, they are parallel. The shortest distance between them is the perpendicular distance from point $P$ to line $l$.
Let $A(-2, 1, -1)$ be the given point on $l$.
$\overrightarrow{AP} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} - \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ 3 \end{pmatrix}$.

3. Calculate the shortest distance using the perpendicular distance formula.

Distance $d = \frac{|\overrightarrow{AP} \times \mathbf{d}|}{|\mathbf{d}|}$
$\overrightarrow{AP} \times \mathbf{d} = \begin{pmatrix} 5 \\ 0 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} (0)(-1) - (3)(0) \\ (3)(1) - (5)(-1) \\ (5)(0) - (0)(1) \end{pmatrix} = \begin{pmatrix} 0 \\ 8 \\ 0 \end{pmatrix}$
$|\overrightarrow{AP} \times \mathbf{d}| = \sqrt{0^2 + 8^2 + 0^2} = 8$
$|\mathbf{d}| = \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2}$
$d = \frac{8}{\sqrt{2}} = 4\sqrt{2}$

1. Express the position vector of $C$ in terms of the parameter $\mu$ and find the displacement vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$.

Since $C$ lies on line $l$, its position vector can be written as:
$\overrightarrow{OC} = \begin{pmatrix} -2 + \mu \\ 1 \\ -1 - \mu \end{pmatrix}$

Find the displacement vectors from $B$ to $A$ and from $B$ to $C$:
$\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \\ -1 \end{pmatrix}$

$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} -2 + \mu \\ 1 \\ -1 - \mu \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 + \mu \\ 0 \\ -1 - \mu \end{pmatrix}$

2. Use the property of perpendicular vectors to solve for $\mu$.

Given that angle $ABC = 90^\circ$, $\overrightarrow{BA}$ and $\overrightarrow{BC}$ are perpendicular, so their scalar (dot) product is zero:
$\overrightarrow{BA} \cdot \overrightarrow{BC} = 0$

$\begin{pmatrix} -3 \\ 0 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} -3 + \mu \\ 0 \\ -1 - \mu \end{pmatrix} = 0$

$-3(-3 + \mu) + 0 + (-1)(-1 - \mu) = 0$
$9 - 3\mu + 1 + \mu = 0$
$10 - 2\mu = 0$
$2\mu = 10 \implies \mu = 5$

3. Substitute $\mu$ back to find the position vector of $C$.

Substitute $\mu = 5$ into the expression for $\overrightarrow{OC}$:
$\overrightarrow{OC} = \begin{pmatrix} -2 + 5 \\ 1 \\ -1 - 5 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ -6 \end{pmatrix}$

The position vector of point $C$ is $3\mathbf{i} + \mathbf{j} - 6\mathbf{k}$.

1. Formulate the equation of the line passing through point $A$ and perpendicular to plane $p_2$.

Let $l_{AF}$ be the line passing through $A$ and perpendicular to $p_2$.
Since it is perpendicular to $p_2$, its direction vector is parallel to the normal vector of $p_2$, which is $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.
The vector equation of line $l_{AF}$ is:
$\mathbf{r} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

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

The equation of $p_2$ is $\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 2$.
At the foot of the perpendicular $F$, the line intersects the plane, so we substitute $\mathbf{r}$ into the plane equation:
$\left[ \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right] \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 2$
$(-2 + 1 - 1) + \lambda(1 + 1 + 1) = 2$
$-2 + 3\lambda = 2$
$3\lambda = 4 \implies \lambda = \frac{4}{3}$

3. Calculate the position vector of the foot of the perpendicular.

Substitute $\lambda = \frac{4}{3}$ back into the equation of the line $l_{AF}$:
$\overrightarrow{OF} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \frac{4}{3} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$
$\overrightarrow{OF} = \begin{pmatrix} -2 + 4/3 \\ 1 + 4/3 \\ -1 + 4/3 \end{pmatrix} = \begin{pmatrix} -2/3 \\ 7/3 \\ 1/3 \end{pmatrix}$

1. Calculate the position vector of the foot of the perpendicular $F$ from $A$ to $p_2$ (result expected from earlier part e).

To solve for $p_3$, we first need the foot of the perpendicular $F$ from $A$ to $p_2$.
The line passing through $A$ perpendicular to $p_2$ has direction vector $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.
Equation of line $AF$: $\mathbf{r} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.
Substituting this into the equation of $p_2$ ($x + y + z = 2$):
$(-2 + \lambda) + (1 + \lambda) + (-1 + \lambda) = 2$
$3\lambda - 2 = 2 \implies \lambda = \frac{4}{3}$
Position vector of $F$ is:
$\overrightarrow{OF} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \frac{4}{3} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -2/3 \\ 7/3 \\ 1/3 \end{pmatrix}$

2. Find the midpoint $M$ of the line segment $AF$.

Since plane $p_3$ is equidistant to point $A$ and plane $p_2$, it must be parallel to $p_2$ and pass exactly midway between $A$ and $p_2$.
Thus, $p_3$ passes through the midpoint $M$ of segment $AF$.
$\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OF}}{2} = \frac{1}{2} \left[ \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \begin{pmatrix} -2/3 \\ 7/3 \\ 1/3 \end{pmatrix} \right] = \begin{pmatrix} -4/3 \\ 5/3 \\ -1/3 \end{pmatrix}$

3. Determine the cartesian equation of plane $p_3$.

Plane $p_3$ is parallel to $p_2$, so it shares the same normal vector $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.
The vector equation of $p_3$ is $\mathbf{r} \cdot \mathbf{n} = \overrightarrow{OM} \cdot \mathbf{n}$:
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -4/3 \\ 5/3 \\ -1/3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$
$x + y + z = -\frac{4}{3} + \frac{5}{3} - \frac{1}{3}$
$x + y + z = 0$