The points $A$ and $B$ have coordinates $(0, 2, 0)$ and $(1, 0, 2)$ respectively, and the plane $\pi_1$ has equation $\mathbf{r} \cdot \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} = 4$.

1. Find the direction vector $\vec{AB}$ from the given coordinates of $A$ and $B$.

$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$

2. Formulate the vector equation of the line passing through $A$ with direction $\vec{AB}$.

$\mathbf{r} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}, \lambda \in \mathbb{R}$

1. Substitute the vector equation of line $AB$ into the equation of plane $\pi_1$.

From part (i), the equation of line $AB$ is $\mathbf{r} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$.
Substitute this into the left hand side of the equation of $\pi_1$:
$\left[ \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} \right] \cdot \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}$

2. Evaluate the scalar product to show it equals 4 for all $\lambda$.

$= \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} + \lambda \left[ \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} \right]$
$= (0 + 4 + 0) + \lambda (2 - 4 + 2)$
$= 4 + \lambda (0)$
$= 4$
Since this equals the right hand side of the plane equation ($4$) for all values of $\lambda$, every point on the line $AB$ satisfies the plane's equation, so the line lies entirely in $\pi_1$.

1. Identify two vectors parallel to plane $\pi_2$.

From part (i), the direction vector of line $AB$ is $\mathbf{d} = \vec{AB} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$.

Since plane $\pi_2$ contains line $AB$, it is parallel to $\mathbf{d}$. Furthermore, because $\pi_2$ is perpendicular to $\pi_1$, it is also parallel to the normal vector of $\pi_1$, which is $\mathbf{n}_1 = \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix}$.

2. Find the normal vector to plane $\pi_2$ using the cross product.

The normal vector $\mathbf{n}_2$ of plane $\pi_2$ must be perpendicular to both $\mathbf{d}$ and $\mathbf{n}_1$:
$\mathbf{n}_2 = \mathbf{d} \times \mathbf{n}_1 = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} \times \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} (-2)(1) - (2)(2) \\ (2)(2) - (1)(1) \\ (1)(2) - (-2)(2) \end{pmatrix} = \begin{pmatrix} -6 \\ 3 \\ 6 \end{pmatrix}$
We can scale this vector by a factor of $-\frac{1}{3}$ to obtain a simpler parallel normal vector $\mathbf{n} = \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}$.

3. Formulate the scalar product equation of plane $\pi_2$.

Using point $A(0, 2, 0)$ which lies on line $AB$ and therefore on $\pi_2$, we apply the scalar product formula $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$:
$\mathbf{r} \cdot \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} = 0(-2) + 2(1) + 0(2) = 2$
Thus, the equation of plane $\pi_2$ is $\mathbf{r} \cdot \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} = 2$ (shown).

1. Express the perpendicular distance from point $P$ to plane $\pi_1$.

The perpendicular distance $d_1$ from $P(p, 0, 0)$ to $\pi_1$ is given by:
$d_1 = \frac{\left| \begin{pmatrix} p \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 2 \\ 1 \end{pmatrix} - 4 \right|}{\sqrt{2^2 + 2^2 + 1^2}} = \frac{|2p - 4|}{3}$

2. Express the perpendicular distance from point $P$ to plane $\pi_2$.

The perpendicular distance $d_2$ from $P(p, 0, 0)$ to $\pi_2$ is given by:
$d_2 = \frac{\left| \begin{pmatrix} p \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} -2 \\ 1 \\ 2 \end{pmatrix} - 2 \right|}{\sqrt{(-2)^2 + 1^2 + 2^2}} = \frac{|-2p - 2|}{3}$

3. Equate the two distances and solve for $p$.

Since the point is equidistant from both planes, $d_1 = d_2$:
$\frac{|2p - 4|}{3} = \frac{|-2p - 2|}{3}$
$|2p - 4| = |-2p - 2|$
$|2p - 4| = |2p + 2|$
Since $2p - 4 = 2p + 2$ has no solution, we consider the negative case:
$2p - 4 = -(2p + 2)$
$4p = 2 \implies p = \frac{1}{2}$

1. Determine the direction of the intersection line and the equidistant point from earlier parts

From part (i), the direction of line $AB$ is:
$\mathbf{d} = \vec{AB} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$

From part (iv), $P(p, 0, 0)$ is equidistant from $\pi_1$ and $\pi_2$:
$\frac{|2p + 2(0) + 1(0) - 4|}{\sqrt{2^2 + 2^2 + 1^2}} = \frac{|-2p + 1(0) + 2(0) - 2|}{\sqrt{(-2)^2 + 1^2 + 2^2}}$
$\frac{|2p - 4|}{3} = \frac{|-2p - 2|}{3}$
$2p - 4 = -2p - 2 \implies 4p = 2 \implies p = \frac{1}{2}$
(Note: $2p - 4 = 2p + 2$ yields no solution).
Thus, $P\left(\frac{1}{2}, 0, 0\right)$ is equidistant from both planes.

2. Deduce the vector equation of the required line

For a line to be equidistant from $\pi_1$ and $\pi_2$ and not intersect either plane, it must be parallel to both planes. This means it must be parallel to their line of intersection.
From (ii) and (iii), both $\pi_1$ and $\pi_2$ contain line $AB$, making $AB$ the unique line of intersection of the two planes.
The required line must therefore be parallel to $AB$ with direction vector $\begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$.
Since the line passes through $P\left(\frac{1}{2}, 0, 0\right)$ which is equidistant from the planes, the equation is:
$\mathbf{r} = \begin{pmatrix} \frac{1}{2} \\ 0 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}, \text{where } \lambda \in \mathbb{R}$