A plane $\Pi_1$ has equation $x + y + z = 3$. A line passes through the points $P$ and $Q$ with position vectors $\mathbf{i} + \mathbf{j} + 2\mathbf{k}$ and $4\mathbf{i} + 2\mathbf{j} - \mathbf{k}$ respectively.

1. Find the displacement vector $\overrightarrow{PQ}$.

$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ -3 \end{pmatrix}$

2. Identify the normal vector to $\Pi_1$ and compute the length of projection of $\overrightarrow{PQ}$ onto this normal.

The cartesian equation of $\Pi_1$ is $x + y + z = 3$, so a normal vector is $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.

The length of projection of $\overrightarrow{PQ}$ onto the normal $\mathbf{n}$ is:
$d = \frac{|\overrightarrow{PQ} \cdot \mathbf{n}|}{|\mathbf{n}|}$
$d = \frac{\left| \begin{pmatrix} 3 \\ 1 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right|}{\sqrt{1^2 + 1^2 + 1^2}} = \frac{|3(1) + 1(1) - 3(1)|}{\sqrt{3}} = \frac{1}{\sqrt{3}}$

3. Use Pythagoras' theorem to find the length of the projection of $\overrightarrow{PQ}$ onto the plane $\Pi_1$.

The magnitude squared of the vector $\overrightarrow{PQ}$ is:
$|\overrightarrow{PQ}|^2 = 3^2 + 1^2 + (-3)^2 = 9 + 1 + 9 = 19$

Let $L$ be the exact length of projection of $\overrightarrow{PQ}$ onto $\Pi_1$. The vector $\overrightarrow{PQ}$, its projection onto the normal, and its projection onto the plane form a right-angled triangle. By Pythagoras' theorem:
$L = \sqrt{|\overrightarrow{PQ}|^2 - d^2}$
$L = \sqrt{19 - \left(\frac{1}{\sqrt{3}}\right)^2} = \sqrt{19 - \frac{1}{3}} = \sqrt{\frac{56}{3}}$

Rationalising the denominator gives:
$L = \frac{\sqrt{56}\sqrt{3}}{3} = \frac{\sqrt{168}}{3} = \frac{\sqrt{4 \times 42}}{3} = \frac{2\sqrt{42}}{3}$

1. Formulate the vector equation of the line $PQ$.

Let $P$ and $Q$ have position vectors $\mathbf{p}$ and $\mathbf{q}$ respectively.
$\overrightarrow{PQ} = \mathbf{q} - \mathbf{p} = \begin{pmatrix} 4 \\ 2 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ -3 \end{pmatrix}$

The vector equation of line $PQ$ is:
$\mathbf{r} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ 1 \\ -3 \end{pmatrix}$, where $\lambda \in \mathbb{R}$.

2. Substitute the line equation into the equation of the plane $\Pi_1$ to find the value of the parameter.

The equation of the plane $\Pi_1$ is $x + y + z = 3$, which can be written as $\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 3$.

At the point of intersection, the position vector satisfies both the line and plane equations:
$\begin{pmatrix} 1 + 3\lambda \\ 1 + \lambda \\ 2 - 3\lambda \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 3$

$(1 + 3\lambda) + (1 + \lambda) + (2 - 3\lambda) = 3$
$4 + \lambda = 3$
$\lambda = -1$

3. Calculate the position vector of the intersection point.

Substitute $\lambda = -1$ back into the line equation:
$\mathbf{r} = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} - 1 \begin{pmatrix} 3 \\ 1 \\ -3 \end{pmatrix} = \begin{pmatrix} -2 \\ 0 \\ 5 \end{pmatrix}$

1. Identify the normal vector for a plane parallel to the $y$-$z$ plane.

A plane parallel to the $y$-$z$ plane is perpendicular to the $x$-axis, so its normal vector is parallel to $\mathbf{i}$.
$\mathbf{n} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

2. Determine the cartesian equation of the plane using the given point.

The equation of the plane is given by $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}$, where $\mathbf{a}$ is the position vector of the point $(-2, 1, 4)$.
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$
$x = -2$

1. Determine the equation of $\Pi_2$ and use it to find $a$

Since $\Pi_2$ is parallel to the $y$-$z$ plane and contains the point $(-2, 1, 4)$, its cartesian equation is $x = -2$.

Given that $S(a, 7, b)$ lies on $\Pi_2$, its $x$-coordinate must satisfy this equation:
$a = -2$

2. Determine $b$ by substituting the coordinates of $S$ into the equation of $\Pi_1$

Since $S(-2, 7, b)$ also lies on $\Pi_1$, its coordinates must satisfy $x + y + z = 3$:
$-2 + 7 + b = 3$
$5 + b = 3$
$b = -2$

1. Find the coordinates of the common point $S$ from the conditions in parts (iii) and (iv).

From (iii), since $\Pi_2$ is parallel to the $y$-$z$ plane and contains $(-2, 1, 4)$, its cartesian equation is $x = -2$.

From (iv), substituting $S(a, 7, b)$ into the equation of $\Pi_2$ gives $a = -2$.
Substituting $S(-2, 7, b)$ into $\Pi_1$ ($x + y + z = 3$):
$(-2) + 7 + b = 3 \implies b = -2$.

Thus, $S$ has position vector $\begin{pmatrix} -2 \\ 7 \\ -2 \end{pmatrix}$ and lies on the line of intersection.

2. Find the direction vector of the line of intersection.

The line of intersection is perpendicular to the normal vectors of both planes.
Let $\mathbf{n}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$.

The direction vector $\mathbf{d}$ is given by their cross product:
$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$.

3. State the vector equation of the line.

Using the point $S$ and the direction vector $\mathbf{d}$, the vector equation of the line is:
$\mathbf{r} = \begin{pmatrix} -2 \\ 7 \\ -2 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}, \lambda \in \mathbb{R}$.