The equations of three planes $p_1, p_2, p_3$ are $2x-4y+z=6,$ $x+y-z=6,$ $ax-2y-z=b,$ respectively, where $a$ and $b$ are constants. The planes $p_1$ and $p_2$ meet in a line $l$.

1. Find the direction vector of the line $l$ by calculating the cross product of the normal vectors of $p_1$ and $p_2$.

The normal vectors of $p_1$ and $p_2$ are $\mathbf{n}_1 = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.

The direction vector of the line $l$ is parallel to $\mathbf{n}_1 \times \mathbf{n}_2$:
$\mathbf{d} = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} (-4)(-1) - (1)(1) \\ (1)(1) - (2)(-1) \\ (2)(1) - (-4)(1) \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 6 \end{pmatrix}$

This direction vector can be scaled down to $\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$.

2. Find the coordinates of a specific point that lies on both planes.

Let $z = 0$. The plane equations then simplify to:
$2x - 4y = 6$
$x + y = 6$

From the second equation, $x = 6 - y$. Substituting this into the first equation:
$2(6 - y) - 4y = 6$
$12 - 2y - 4y = 6$
$6y = 6 \implies y = 1$
Then $x = 6 - 1 = 5$.

A common point on both planes (and therefore on $l$) is $(5, 1, 0)$.

3. State the vector equation of the line $l$.

Using the point $(5, 1, 0)$ and the simplified direction vector $\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$:
$\mathbf{r} = \begin{pmatrix} 5 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}, \lambda \in \mathbb{R}$

1. Reduce the system of equations by eliminating $x$ from the second and third equations.

Let the equations be:
(1) $x + y - z = 6$
(2) $2x - 4y + z = 6$
(3) $ax - 2y - z = b$

We eliminate $x$ from equations (2) and (3) using equation (1).
Take (2) $- 2 \times$ (1):
$(2x - 4y + z) - 2(x + y - z) = 6 - 2(6)$
$-6y + 3z = -6 \quad \text{--- (4)}$

Take (3) $- a \times$ (1):
$(ax - 2y - z) - a(x + y - z) = b - 6a$
$(-2-a)y + (a-1)z = b - 6a \quad \text{--- (5)}$

2. Apply the condition for infinite solutions to find $a$.

For the 3x3 system to have an infinite number of solutions (intersecting in a line), the remaining equations after eliminating one variable must be dependent. This means equation (5) must be proportional to equation (4).

By comparing the ratio of their coefficients:
$\frac{-2-a}{-6} = \frac{a-1}{3} = \frac{b-6a}{-6}$

Using the first equality to solve for $a$:
$\frac{-2-a}{-6} = \frac{a-1}{3}$
$-2 - a = -2(a - 1)$
$-2 - a = -2a + 2$
$a = 4$

3. Use the value of $a$ and the proportional relationship to find $b$.

Substitute $a = 4$ into the second equality to solve for $b$:
$\frac{4-1}{3} = \frac{b-6(4)}{-6}$
$\frac{3}{3} = \frac{b - 24}{-6}$
$1 = \frac{b - 24}{-6}$
$-6 = b - 24$
$b = 18$

1. Find the normal vector to the plane by calculating the cross product of the normal vectors of $p_1$ and $p_2$.

Let the normal vectors of $p_1$ and $p_2$ be $\mathbf{n}_1$ and $\mathbf{n}_2$ respectively.
$\mathbf{n}_1 = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}, \quad \mathbf{n}_2 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$
Since the required plane is perpendicular to both $p_1$ and $p_2$, its normal vector $\mathbf{n}$ must be perpendicular to both $\mathbf{n}_1$ and $\mathbf{n}_2$, meaning it is parallel to their cross product.
$\mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} (-4)(-1) - (1)(1) \\ (1)(1) - (2)(-1) \\ (2)(1) - (-4)(1) \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ 6 \end{pmatrix}$
We can factor out $3$ to choose a simpler normal vector for our plane, so let $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$.

2. Use the given point to find the scalar constant $p$ and write the final equation of the plane.

The required plane passes through the point with position vector $\mathbf{a} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.
The equation of the plane in scalar product form is $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$.
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = (1)(1) + (1)(1) + (1)(2) = 4$

1. Deduce the geometrical relationship between lines $l$ and $m$.

Since $l$ is the intersection of $p_1$ and $p_2$, and $m$ is the intersection of $p_4$ and $p_2$, both lines $l$ and $m$ lie completely in the plane $p_2$.
If $l$ and $m$ were to intersect at a point, this point would lie in $p_1$, $p_2$, and $p_4$. However, we are given that $p_1, p_2,$ and $p_4$ have no point in common.
Therefore, $l$ and $m$ cannot intersect. Since they are coplanar and non-intersecting, they must be parallel lines ($l \parallel m$).

2. Find a displacement vector between a point on $l$ and the given point on $m$.

From part (i), the equation of line $l$ is $\mathbf{r} = \begin{pmatrix} 5 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$.
Let $B(5, 1, 0)$ be a point on $l$. The direction vector of $l$ (and therefore $m$) is $\mathbf{d} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}$.
Let $A(3, 4, 1)$ be the given point on $m$.
The displacement vector from $B$ to $A$ is:
$\vec{BA} = \begin{pmatrix} 3 \\ 4 \\ 1 \end{pmatrix} - \begin{pmatrix} 5 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}$

3. Calculate the perpendicular distance between the parallel lines $l$ and $m$.

The perpendicular distance $d$ from point $A$ to line $l$ is given by $d = \frac{|\vec{BA} \times \mathbf{d}|}{|\mathbf{d}|}$.
$\vec{BA} \times \mathbf{d} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} (3)(2) - (1)(1) \\ (1)(1) - (-2)(2) \\ (-2)(1) - (3)(1) \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \\ -5 \end{pmatrix}$
$d = \frac{\sqrt{5^2 + 5^2 + (-5)^2}}{\sqrt{1^2 + 1^2 + 2^2}} = \frac{\sqrt{75}}{\sqrt{6}} = \frac{5\sqrt{3}}{\sqrt{6}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2}$