The equations of two planes $P_1$ and $P_2$ are $x - 2y + 3z = 4$ and $3x + 2y - z = 4$ respectively. The equation of a third plane $P_3$ is $5x - ky + 6z = 1$, where $k$ is a constant.

1. Find the direction vector of the line of intersection by taking the cross product of the normal vectors of the two planes.

The normal vectors to the planes are $\mathbf{n}_1 = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}$.
The line $L$ lies on both planes, so its direction vector $\mathbf{d}$ is perpendicular to both normal vectors:

$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} \times \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}$
$= \begin{pmatrix} (-2)(-1) - (3)(2) \\ (3)(3) - (1)(-1) \\ (1)(2) - (-2)(3) \end{pmatrix}$
$= \begin{pmatrix} -4 \\ 10 \\ 8 \end{pmatrix} = 2 \begin{pmatrix} -2 \\ 5 \\ 4 \end{pmatrix}$

We can take the simplest parallel vector $\begin{pmatrix} -2 \\ 5 \\ 4 \end{pmatrix}$ as the direction vector of $L$.

2. Find a particular point that lies on both planes.

To find a point on $L$, we need to find a coordinate that satisfies both plane equations.
Let $z = 0$:
$x - 2y = 4$ --- (1)
$3x + 2y = 4$ --- (2)

Adding (1) and (2):
$4x = 8 \implies x = 2$

Substituting $x = 2$ into (1):
$2 - 2y = 4 \implies 2y = -2 \implies y = -1$

Thus, a point on the line is $(2, -1, 0)$.

3. Write the vector equation of the line $L$.

Combining the position vector of the point and the direction vector, the vector equation of $L$ is:
$\mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ 5 \\ 4 \end{pmatrix}$, where $\lambda \in \mathbb{R}$.

1. State the direction vector of line L (from part (i)).

The direction of $L$ is $\mathbf{n_1}\times\mathbf{n_2}$ where $\mathbf{n_1}=\begin{pmatrix}1\\-2\\3\end{pmatrix}$, $\mathbf{n_2}=\begin{pmatrix}3\\2\\-1\end{pmatrix}$.

$\mathbf{n_1}\times\mathbf{n_2}=\begin{pmatrix}(-2)(-1)-(3)(2)\\(3)(3)-(1)(-1)\\(1)(2)-(-2)(3)\end{pmatrix}=\begin{pmatrix}-4\\10\\8\end{pmatrix}=2\begin{pmatrix}-2\\5\\4\end{pmatrix}.$

So $L$ has direction $\mathbf{d}=\begin{pmatrix}-2\\5\\4\end{pmatrix}$.

2. Apply the 'no common point' condition: P3 parallel to L, and solve for k.

For the three planes to have no point in common, $P_3$ must be parallel to $L$, so the normal of $P_3$, $\mathbf{n_3}=\begin{pmatrix}5\\-k\\6\end{pmatrix}$, is perpendicular to $\mathbf{d}$:

$\mathbf{n_3}\cdot\mathbf{d}=0$
$5(-2)+(-k)(5)+6(4)=0$
$-10-5k+24=0\;\Rightarrow\;14-5k=0\;\Rightarrow\;k=\frac{14}{5}.$

(Check: with the point $(2,-1,0)$ on $L$, $5(2)-\tfrac{14}{5}(-1)+6(0)=12.8\neq 1$, so $P_3$ does not contain $L$ — there is genuinely no common point.)

1. Find the coordinates of Q where L meets the x-y plane

From part (i), $L:\ \mathbf{r} = \begin{pmatrix}2\\-1\\0\end{pmatrix} + t\begin{pmatrix}-2\\5\\4\end{pmatrix}$.

The x-y plane is $z = 0$, so $0 + 4t = 0 \Rightarrow t = 0$.

Hence $Q = (2,\,-1,\,0)$.

2. State P_3 using the value of k from part (ii)

From part (ii), $k = \tfrac{14}{5}$, so $P_3:\ 5x - \tfrac{14}{5}y + 6z = 1$.

Multiplying through by $5$: $\ 25x - 14y + 30z = 5$, with normal $\mathbf{n} = \begin{pmatrix}25\\-14\\30\end{pmatrix}$.

3. Apply the point-to-plane distance formula

Distance $= \dfrac{|25(2) - 14(-1) + 30(0) - 5|}{\sqrt{25^2 + 14^2 + 30^2}}$

$= \dfrac{|50 + 14 - 5|}{\sqrt{625 + 196 + 900}} = \dfrac{59}{\sqrt{1721}} \approx 1.42$ units.

1. Recall or compute the coordinates of $Q$ and the value of $k$ from previous parts.

For $Q$, the line $L$ meets the $x$-$y$ plane where $z=0$. Substituting $z=0$ into the equations of $P_1$ and $P_2$:
$x - 2y = 4$
$3x + 2y = 4$
Solving simultaneously gives $4x = 8 \implies x=2$ and $y=-1$. Thus, $Q$ is $(2, -1, 0)$.

For $k$, the three planes having no point in common means the intersection line $L$ is parallel to $P_3$. Using the direction vector of $L$, $\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} \times \begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} -4 \\ 10 \\ 8 \end{pmatrix}$:
$\mathbf{d} \cdot \mathbf{n}_3 = 0 \implies \begin{pmatrix} -4 \\ 10 \\ 8 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ -k \\ 6 \end{pmatrix} = 0 \implies -20 - 10k + 48 = 0 \implies k = 2.8$

2. Find the equation of the plane containing $Q$ and parallel to $P_3$.

Since the new plane is parallel to $P_3$, it shares the same normal vector $\mathbf{n} = \begin{pmatrix} 5 \\ -2.8 \\ 6 \end{pmatrix}$.
The equation of the plane passing through $Q(2, -1, 0)$ is:
$\mathbf{r} \cdot \begin{pmatrix} 5 \\ -2.8 \\ 6 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 5 \\ -2.8 \\ 6 \end{pmatrix}$
$5x - 2.8y + 6z = 5(2) - 2.8(-1) + 6(0)$
$5x - 2.8y + 6z = 12.8$

3. Determine the relative positions of the origin $O$ and $Q$ with respect to $P_3$.

The equation of $P_3$ is $5x - 2.8y + 6z = 1$.
The parallel plane containing the origin $O(0,0,0)$ is $5x - 2.8y + 6z = 0$.

Comparing the constants on the RHS for these three parallel planes:
$0 < 1 < 12.8$

Since the constant for $P_3$ ($1$) lies strictly between the constants for the planes containing $O$ ($0$) and $Q$ ($12.8$), $P_3$ lies spatially between $O$ and $Q$.
Therefore, $O$ and $Q$ are on opposite sides of $P_3$.