The equations of four planes $p_{1}$, $p_{2}$, $p_{3}$ and $p_{4}$ are $x+y+z=2,\\-x+ay+z=1,\\2x+2y+2z=b,\\cx+2y+3z=d,$ respectively, where $a$, $b$, $c$ and $d$ are constants. The planes $p_{1}$ and $p_{2}$ intersect in a line l whose equation is $x-1=\dfrac{y-1}{2}= -\dfrac{z}{3}$ .

1. Identify the direction vector of the line and the normal vector of the plane $p_2$.

From the given equation of line $l$, $x-1=\dfrac{y-1}{2}= \dfrac{z}{-3}$, the direction vector of the line is $\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$.

From the Cartesian equation of plane $p_{2}$, $-x+ay+z=1$, the normal vector to the plane is $\mathbf{n}_2 = \begin{pmatrix} -1 \\ a \\ 1 \end{pmatrix}$.

2. Use the orthogonality of the line's direction and the plane's normal to solve for $a$.

Since the line $l$ is the intersection of $p_1$ and $p_2$, it lies entirely on $p_2$. Therefore, its direction vector must be perpendicular to the normal vector of $p_{2}$.
$\mathbf{d} \cdot \mathbf{n}_2 = 0$
$\begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ a \\ 1 \end{pmatrix} = 0$
$(1)(-1) + (2)(a) + (-3)(1) = 0$
$-1 + 2a - 3 = 0$
$2a = 4$
$a = 2 \text{ (shown)}$

1. Formulate the equation of the line passing through $A$ perpendicular to $p_1$.

The normal vector to the plane $p_1: x+y+z=2$ is $\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$.

The line passing through $A(0,0,1)$ perpendicular to $p_1$ has the equation:
$\mathbf{r} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

2. Find the value of the parameter $\lambda$ at the intersection point.

Since the foot of the perpendicular $N$ lies on $p_1$, its position vector satisfies the plane equation $\mathbf{r} \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 2$:
$\left[ \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right] \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 2$
$(0)(1) + (0)(1) + (1)(1) + \lambda(1^2 + 1^2 + 1^2) = 2$
$1 + 3\lambda = 2$
$3\lambda = 1 \implies \lambda = \frac{1}{3}$

3. Calculate the position vector of the foot of the perpendicular.

Substitute $\lambda = \frac{1}{3}$ into the line equation to find the position vector of $N$:
$\vec{ON} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} + \frac{1}{3} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ \frac{1}{3} \\ \frac{4}{3} \end{pmatrix}$

1. Identify the normal vectors of planes $p_1$ and $p_2$.

From the equations $x+y+z=2$ and $-x+2y+z=1$ (using $a=2$ from part i), the normal vectors are:
$\mathbf{n}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}$.

2. Set up the scalar product equation to find the angle between the planes.

Let $\theta$ be the acute angle between the planes. We use the formula $\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1| |\mathbf{n}_2|}$:
$\cos \theta = \frac{|(1)(-1) + (1)(2) + (1)(1)|}{\sqrt{1^2+1^2+1^2} \sqrt{(-1)^2+2^2+1^2}}$
$\cos \theta = \frac{|-1+2+1|}{\sqrt{3}\sqrt{6}}$

3. Calculate the acute angle.

$\cos \theta = \frac{2}{\sqrt{18}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3}$
$\theta = \cos^{-1}\left(\frac{\sqrt{2}}{3}\right) \approx 61.87^\circ \approx 61.9^\circ$ (to 1 d.p.)

1. Identify a point on $p_{1}$ and the normal vector of $p_{3}$.

The equation of $p_{1}$ is $x+y+z=2$.
Let $y=0$ and $z=0$, then $x=2$.
A point on $p_{1}$ is $(2, 0, 0)$, which has position vector $\mathbf{a} = \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}$.

The equation of $p_{3}$ is $2x+2y+2z=b$, which can be written as $\mathbf{r} \cdot \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} = b$.

2. Set up the distance equation from the point on $p_{1}$ to $p_{3}$.

The perpendicular distance from point $\mathbf{a}$ to a plane $\mathbf{r} \cdot \mathbf{n} = D$ is given by $\frac{|\mathbf{a} \cdot \mathbf{n} - D|}{|\mathbf{n}|}$.

Distance $= \frac{\left| \begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} - b \right|}{\sqrt{2^2 + 2^2 + 2^2}} = \frac{|4 - b|}{\sqrt{12}} = \frac{|4 - b|}{2\sqrt{3}}$

Given the distance is $\sqrt{3}$:
$\frac{|4 - b|}{2\sqrt{3}} = \sqrt{3}$

3. Solve the modulus equation for $b$.

$|4 - b| = (\sqrt{3})(2\sqrt{3})$
$|4 - b| = 6$

$4 - b = 6 \quad \text{or} \quad 4 - b = -6$
$b = -2 \quad \text{or} \quad b = 10$

Since it is given that $b > 0$, we reject $b = -2$.
Therefore, $b = 10$.

1. Deduce that plane $p_4$ must contain the line $l$.

Since $p_1$ and $p_2$ intersect exactly at the line $l$, for $p_1$, $p_2$ and $p_4$ to meet in a line, the plane $p_4$ must contain the intersection line of $p_1$ and $p_2$, which is $l$.

2. Use the orthogonality of the line's direction vector and the plane's normal vector to find $c$.

The direction vector of $l$ is $\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$ and the normal vector to $p_4$ is $\mathbf{n}_4 = \begin{pmatrix} c \\ 2 \\ 3 \end{pmatrix}$.
Since $l$ lies on $p_4$, its direction vector is perpendicular to the normal of $p_4$, meaning their dot product is zero: $\mathbf{d} \cdot \mathbf{n}_4 = 0$.
$\begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} c \\ 2 \\ 3 \end{pmatrix} = 0$
$c + 4 - 9 = 0 \implies c = 5$

3. Substitute a point on the line into the equation of $p_4$ to find $d$.

From the Cartesian equation of $l$, a point on the line is $(1, 1, 0)$.
Since $l$ lies on $p_4$, this point must satisfy the equation of $p_4$ ($5x + 2y + 3z = d$).
$5(1) + 2(1) + 3(0) = d$
$d = 7$

1. State the geometric condition for the three planes to have no common point.

We are given that $p_1$ and $p_2$ intersect at the line $l$. For the planes $p_1$, $p_2$ and $p_4$ to have no point in common, $l$ must be parallel to $p_4$ but must not lie on it.

2. Find the value of $c$ for $l$ to be parallel to $p_4$.

The direction vector of $l$ is $\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$ and the normal vector to $p_4$ is $\mathbf{n}_4 = \begin{pmatrix} c \\ 2 \\ 3 \end{pmatrix}$.
For $l$ to be parallel to $p_4$, $\mathbf{d}$ must be perpendicular to $\mathbf{n}_4$:
$\mathbf{d} \cdot \mathbf{n}_4 = 0$
$\begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} c \\ 2 \\ 3 \end{pmatrix} = 0$
$c + 4 - 9 = 0 \implies c = 5$

3. Determine the restriction on $d$ so that $l$ does not lie on $p_4$.

From the equation of $l$, a point on the line is $(1, 1, 0)$.
For $l$ to not lie entirely on $p_4$, this point must not satisfy the equation of $p_4$:
$c(1) + 2(1) + 3(0) \neq d$
Substituting $c = 5$:
$5(1) + 2 \neq d \implies d \neq 7$
(Note: This can also be directly deduced from part (v), where $c=5, d=7$ yields an infinite number of common points on $l$.)