The plane $p$ has equation $2x - 3y + 6z = 16$. The plane $q$ is parallel to $p$ and contains the point with position vector $\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}$.

1. State the general form of the equation for a plane parallel to $p$.

Since plane $q$ is parallel to plane $p$: $2x - 3y + 6z = 16$, it shares the same normal vector. The equation of $q$ can therefore be written in the form:
$2x - 3y + 6z = d$

2. Substitute the coordinates of the given point to find $d$ and state the final equation.

Plane $q$ contains the point with position vector $\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}$, which corresponds to the coordinates $(1, 4, 2)$. Substituting these coordinates into the general equation gives:
$2(1) - 3(4) + 6(2) = d$
$d = 2 - 12 + 12 = 2$
Therefore, the equation of plane $q$ is $2x - 3y + 6z = 2$.

1. Set up the perpendicular distance formula using the given position vector of a point on plane $q$ and the normal vector of plane $p$.

$\text{Distance} = \frac{\left| \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix} - 16 \right|}{\left| \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix} \right|}$

2. Evaluate the scalar product and the magnitude of the normal vector to find the exact distance.

$= \frac{|2(1) - 3(4) + 6(2) - 16|}{\sqrt{2^2 + (-3)^2 + 6^2}}$
$= \frac{|2 - 12 + 12 - 16|}{\sqrt{4 + 9 + 36}}$
$= \frac{|-14|}{\sqrt{49}}$
$= \frac{14}{7} = 2$

1. Identify the normal vectors of the two given planes.

Let the normal vectors of plane $p$ and the second plane be $\mathbf{n}_1$ and $\mathbf{n}_2$ respectively.
From their Cartesian equations $2x - 3y + 6z = 16$ and $x - 2y + 2z = 5$, we deduce:
$\mathbf{n}_1 = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$.

2. Determine the relationship between the direction vector of the line and the normal vectors of the planes.

Since line $l$ is parallel to both planes, its direction vector $\mathbf{d}$ must be perpendicular to both normal vectors $\mathbf{n}_1$ and $\mathbf{n}_2$.
Therefore, $\mathbf{d}$ is parallel to the cross product $\mathbf{n}_1 \times \mathbf{n}_2$.

3. Calculate the cross product to find the direction vector of the line.

$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix} \times \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = \begin{pmatrix} (-3)(2) - (6)(-2) \\ (6)(1) - (2)(2) \\ (2)(-2) - (-3)(1) \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \\ -1 \end{pmatrix}$

4. Formulate the vector equation of the line $l$.

Since line $l$ passes through the origin, its position vector is $\mathbf{0}$.
The vector equation of $l$ is $\mathbf{r} = \mathbf{0} + \lambda \mathbf{d}$, which gives:
$\mathbf{r} = \lambda \begin{pmatrix} 6 \\ 2 \\ -1 \end{pmatrix}, \quad \lambda \in \mathbb{R}$