The plane $p$ has equation $x - 2y + 2z = 12$. With reference to the origin $O$, the point $A$ has position vector $2\mathbf{i} + \mathbf{j} - 3\mathbf{k}$.

1. Write the equation of the line passing through $A$ and perpendicular to the plane $p$.

The normal vector of the plane $p$ is $\mathbf{n} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$.
The equation of the line passing through $A$ and perpendicular to $p$ is
$\mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}, \text{ where } \lambda \in \mathbb{R}.$

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

Since $N$ lies on the plane $p$, its position vector satisfies the plane equation $\mathbf{r} \cdot \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = 12$:
$\left[ \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} \right] \cdot \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = 12$
$(2 + \lambda) - 2(1 - 2\lambda) + 2(-3 + 2\lambda) = 12$
$2 + \lambda - 2 + 4\lambda - 6 + 4\lambda = 12$
$9\lambda - 6 = 12 \implies 9\lambda = 18 \implies \lambda = 2$

3. Evaluate the position vector of $N$ and state its coordinates.

Substitute $\lambda = 2$ back into the line equation to find $\vec{ON}$:
$\vec{ON} = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix}$
The coordinates of $N$ are $(4, -3, 1)$.

4. Use the midpoint property to find the coordinates of the reflection $A'$.

Let the reflection of $A$ in $p$ be $A'$. The foot of perpendicular $N$ is the midpoint of the line segment $AA'$.
$\vec{ON} = \frac{\vec{OA} + \vec{OA'}}{2}$
$\vec{OA'} = 2\vec{ON} - \vec{OA}$
$\vec{OA'} = 2 \begin{pmatrix} 4 \\ -3 \\ 1 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 8 \\ -6 \\ 2 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 6 \\ -7 \\ 5 \end{pmatrix}$
The coordinates of the reflection of $A$ in $p$ are $(6, -7, 5)$.

1. Set up the general equation for planes parallel to $p$.

Since the required planes are parallel to $p$, they share the same normal vector $\mathbf{n} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$.
Let the equation of the required planes be $x - 2y + 2z = d$.

2. Express the perpendicular distance using a specific point on $p$.

By substituting $y = 0$ and $z = 0$ into the equation of $p$ ($x - 2y + 2z = 12$), we find that $(12, 0, 0)$ is a point on $p$.
The perpendicular distance from $(12, 0, 0)$ to the plane $x - 2y + 2z = d$ is $15$.
Using the perpendicular distance formula:
$\frac{|1(12) - 2(0) + 2(0) - d|}{\sqrt{1^2 + (-2)^2 + 2^2}} = 15$
$\frac{|12 - d|}{\sqrt{9}} = 15$
$\frac{|12 - d|}{3} = 15$

3. Solve for $d$ and state the final cartesian equations.

$|12 - d| = 45$
$12 - d = 45 \quad \text{or} \quad 12 - d = -45$
$d = -33 \quad \text{or} \quad d = 57$

The cartesian equations of the planes are $x - 2y + 2z = -33$ and $x - 2y + 2z = 57$.