平面 $p_1$ 的方程为 $x - y + z = 0$，平面 $p_2$ 的方程为 $x + y + z = 2$。

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

From the equations $x - y + z = 0$ and $x + y + z = 2$, the normal vectors to planes $p_1$ and $p_2$ are respectively:
$\mathbf{n}_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

2. Use the scalar product to find the acute angle between the normal vectors.

Let $\theta$ be the acute angle between the planes.
$\cos \theta = \frac{| \mathbf{n}_1 \cdot \mathbf{n}_2 |}{|\mathbf{n}_1| |\mathbf{n}_2|}$
$\cos \theta = \frac{|(1)(1) + (-1)(1) + (1)(1)|}{\sqrt{1^2+(-1)^2+1^2} \sqrt{1^2+1^2+1^2}}$
$\cos \theta = \frac{1}{\sqrt{3}\sqrt{3}} = \frac{1}{3}$
$\theta = \cos^{-1}\left(\frac{1}{3}\right) \approx 70.5287^\circ \approx 70.5^\circ \text{ (1 d.p.)}$

1. Find the direction vector of the line of intersection using the cross product of the planes' normal vectors.

$\mathbf{n}_1 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \quad \mathbf{n}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

The direction vector $\mathbf{d}$ of the line of intersection is given by $\mathbf{n}_1 \times \mathbf{n}_2$:
$\mathbf{d} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} (-1)(1) - (1)(1) \\ (1)(1) - (1)(1) \\ (1)(1) - (-1)(1) \end{pmatrix} = \begin{pmatrix} -2 \\ 0 \\ 2 \end{pmatrix}$

Taking a simpler parallel vector (by dividing by 2), we use $\mathbf{d} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}$.

2. Find a point that lies on both planes to serve as the position vector for the line.

Let $z = 0$. The Cartesian equations of the planes become:
$x - y = 0 \implies x = y$
$x + y = 2$

Substituting $x = y$ into the second equation:
$2y = 2 \implies y = 1 \text{ and } x = 1$

A common point on the line is $(1, 1, 0)$, so the position vector is $\mathbf{a} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$.

3. State the vector equation of the line.

The equation of the line is $\mathbf{r} = \mathbf{a} + \lambda \mathbf{d}$, where $\lambda \in \mathbb{R}$.
$\mathbf{r} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \quad \lambda \in \mathbb{R}$

1. Express the equation of the second line in vector form and observe it is parallel to $l$.

Let the second line be $l_2$. Its cartesian equation is given by $x - 3 = \frac{z - 2}{-1}$, $y = 1$.
By converting this to vector form, a point on $l_2$ is $P(3, 1, 2)$ and its direction vector is $\mathbf{d}_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$.
The line $l$ has equation $\mathbf{r} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}$.
Since the direction vectors are identical, line $l$ and line $l_2$ are parallel.

2. Calculate the displacement vector between points on the two lines and its cross product with the direction vector.

Let $A(-2, 1, -1)$ be the point on $l$ corresponding to position vector $\overrightarrow{OA}$.
The displacement vector from $A$ to $P$ is:
$\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} - \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 5 \\ 0 \\ 3 \end{pmatrix}$
The cross product of $\overrightarrow{AP}$ with the direction vector $\mathbf{d} = \mathbf{i} - \mathbf{k}$ is:
$\overrightarrow{AP} \times \mathbf{d} = \begin{pmatrix} 5 \\ 0 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} (0)(-1) - (3)(0) \\ (3)(1) - (5)(-1) \\ (5)(0) - (0)(1) \end{pmatrix} = \begin{pmatrix} 0 \\ 8 \\ 0 \end{pmatrix}$

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

The shortest distance between the two parallel lines is the perpendicular distance from point $P$ on $l_2$ to line $l$:
$\text{Distance} = \frac{|\overrightarrow{AP} \times \mathbf{d}|}{|\mathbf{d}|} = \frac{\sqrt{0^2 + 8^2 + 0^2}}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}$

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

Since $C$ lies on $l$, its position vector $\overrightarrow{OC}$ is:
$\overrightarrow{OC} = \begin{pmatrix} -2 + \mu \\ 1 \\ -1 - \mu \end{pmatrix}$
Next, we find the displacement vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$:
$\overrightarrow{BA} = \overrightarrow{OA} - \overrightarrow{OB} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 \\ 0 \\ -1 \end{pmatrix}$
$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} -2 + \mu \\ 1 \\ -1 - \mu \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -3 + \mu \\ 0 \\ -1 - \mu \end{pmatrix}$

2. Set up the scalar product equation for the perpendicular vectors and solve for $\mu$.

Since angle $ABC = 90^\circ$, the vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$ are perpendicular, which means their scalar product is zero:
$\overrightarrow{BA} \cdot \overrightarrow{BC} = 0$
$\begin{pmatrix} -3 \\ 0 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} -3 + \mu \\ 0 \\ -1 - \mu \end{pmatrix} = 0$
$-3(-3 + \mu) + 0(0) - 1(-1 - \mu) = 0$
$9 - 3\mu + 1 + \mu = 0$
$10 - 2\mu = 0 \implies \mu = 5$

3. Substitute $\mu$ back into $\overrightarrow{OC}$ to find the position vector of point $C$.

Substitute $\mu = 5$ into the expression for $\overrightarrow{OC}$:
$\overrightarrow{OC} = \begin{pmatrix} -2 + 5 \\ 1 \\ -1 - 5 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ -6 \end{pmatrix}$
So, the position vector of point $C$ is $3\mathbf{i} + \mathbf{j} - 6\mathbf{k}$.

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

Let the line be $l_{AF}$. Since it is perpendicular to $p_2$, its direction vector is the normal vector of $p_2$, which is $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.

The vector equation of the line is:
$l_{AF}: \mathbf{r} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \text{ where } \lambda \in \mathbb{R}$

2. Substitute the line equation into the plane equation to find the value of the parameter $\lambda$ at the intersection.

The equation of $p_2$ is $\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 2$. Substitute $l_{AF}$ into $p_2$:
$\left[ \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right] \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 2$
$\begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 2$
$(-2 + 1 - 1) + \lambda(1 + 1 + 1) = 2$
$-2 + 3\lambda = 2$
$3\lambda = 4 \implies \lambda = \frac{4}{3}$

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

Substitute $\lambda = \frac{4}{3}$ back into the equation of the line $l_{AF}$ to find the position vector of the foot of the perpendicular, $F$:
$\overrightarrow{OF} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \frac{4}{3} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 + 4/3 \\ 1 + 4/3 \\ -1 + 4/3 \end{pmatrix} = \begin{pmatrix} -2/3 \\ 7/3 \\ 1/3 \end{pmatrix}$

Expressing in terms of $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$:
$\overrightarrow{OF} = -\frac{2}{3}\mathbf{i} + \frac{7}{3}\mathbf{j} + \frac{1}{3}\mathbf{k}$

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

Let the normal to plane $p_2$ be $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.
The equation of the line passing through $A$ and perpendicular to $p_2$ is:
$\mathbf{r} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$
Substitute this into the equation of $p_2$ ($x + y + z = 2$) to find the intersection:
$(-2 + \lambda) + (1 + \lambda) + (-1 + \lambda) = 2$
$3\lambda - 2 = 2 \implies \lambda = \frac{4}{3}$
The position vector of the foot of perpendicular $F$ is:
$\overrightarrow{OF} = \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \frac{4}{3} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -2/3 \\ 7/3 \\ 1/3 \end{pmatrix}$
(Note: This calculation is typically completed in part (e) and can be carried forward directly).

2. Determine the equation of $p_3$ by finding the midpoint between $A$ and $F$.

For plane $p_3$ to be equidistant to point $A$ and plane $p_2$ with a strictly non-zero distance, it must be parallel to $p_2$ and pass through the midpoint $M$ of $A$ and $F$.
The position vector of $M$ is:
$\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OF}}{2} = \frac{1}{2} \left[ \begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix} + \begin{pmatrix} -2/3 \\ 7/3 \\ 1/3 \end{pmatrix} \right] = \begin{pmatrix} -4/3 \\ 5/3 \\ -1/3 \end{pmatrix}$
Since $p_3$ is parallel to $p_2$, its normal vector is also $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.
The scalar product equation of $p_3$ is $\mathbf{r} \cdot \mathbf{n} = \overrightarrow{OM} \cdot \mathbf{n}$:
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -4/3 \\ 5/3 \\ -1/3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = -\frac{4}{3} + \frac{5}{3} - \frac{1}{3} = 0$
Thus, the cartesian equation of $p_3$ is $x + y + z = 0$.