The plane $\Pi_1$ contains the points $A(1,4,2),\ B(1,0,5)$ and $C(0,8,-1)$.

1. Find two direction vectors lying on the plane $\Pi_1$.

$\overrightarrow{AB} = \begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -4 \\ 3 \end{pmatrix}$

$\overrightarrow{AC} = \begin{pmatrix} 0 \\ 8 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ -3 \end{pmatrix}$

2. Calculate the normal vector to the plane using the cross product of the two direction vectors.

$\mathbf{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{pmatrix} 0 \\ -4 \\ 3 \end{pmatrix} \times \begin{pmatrix} -1 \\ 4 \\ -3 \end{pmatrix} = \begin{pmatrix} (-4)(-3) - (3)(4) \\ (3)(-1) - (0)(-3) \\ (0)(4) - (-4)(-1) \end{pmatrix}$

$\mathbf{n} = \begin{pmatrix} 0 \\ -3 \\ -4 \end{pmatrix}$

Taking a parallel normal vector with positive components, let $\mathbf{n}' = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix}$.

3. Formulate the cartesian equation of the plane using the normal vector and point $A$.

The vector equation of the plane is given by $\mathbf{r} \cdot \mathbf{n}' = \mathbf{a} \cdot \mathbf{n}'$.

$\mathbf{r} \cdot \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix}$

$\mathbf{r} \cdot \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} = 0(1) + 3(4) + 4(2) = 20$

Substituting $\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ gives:
$\begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} = 20 \implies 3y + 4z = 20$

1. Determine the cartesian equation of the plane $\Pi_1$ (required from part a).

From part (a), we first need to determine the normal vector and the cartesian equation of $\Pi_1$. Using points $A(1,4,2)$, $B(1,0,5)$ and $C(0,8,-1)$:
$\vec{AB} = \begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix} - \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ -4 \\ 3 \end{pmatrix}$
$\vec{AC} = \begin{pmatrix} 0 \\ 8 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ -3 \end{pmatrix}$
A normal vector to $\Pi_1$ is $\mathbf{n} = \vec{AB} \times \vec{AC} = \begin{pmatrix} 0 \\ -3 \\ -4 \end{pmatrix}$.
Taking a parallel vector $\mathbf{n} = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix}$, the equation of $\Pi_1$ is:
$\mathbf{r} \cdot \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} = 0(1) + 3(4) + 4(2) = 20$
So the cartesian equation is $3y + 4z = 20$.

2. Calculate the perpendicular distance from $P$ to $\Pi_1$.

Using the perpendicular distance formula from a point to a plane, $d = \frac{|ax_0+by_0+cz_0-D|}{\sqrt{a^2+b^2+c^2}}$, we substitute the coordinates of $P(2, 1, 2)$:
$\text{Distance} = \frac{|0(2) + 3(1) + 4(2) - 20|}{\sqrt{0^2 + 3^2 + 4^2}}$
$= \frac{|3 + 8 - 20|}{5}$
$= \frac{|-9|}{5} = \frac{9}{5}$

1. Derive the cartesian equation of plane $\Pi_1$ (required from part a).

To find the normal to $\Pi_1$, we use vectors $\vec{AB}$ and $\vec{AC}$:
$\vec{AB} = \begin{pmatrix} 1-1 \\ 0-4 \\ 5-2 \end{pmatrix} = \begin{pmatrix} 0 \\ -4 \\ 3 \end{pmatrix}$
$\vec{AC} = \begin{pmatrix} 0-1 \\ 8-4 \\ -1-2 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ -3 \end{pmatrix}$

$\mathbf{n}_1 = \vec{AB} \times \vec{AC} = \begin{pmatrix} 0 \\ -4 \\ 3 \end{pmatrix} \times \begin{pmatrix} -1 \\ 4 \\ -3 \end{pmatrix} = \begin{pmatrix} 0 \\ -3 \\ -4 \end{pmatrix}$
Taking a parallel normal vector $\begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix}$, the equation is of the form $3y + 4z = k$.
Substitute point $A(1,4,2)$:
$k = 3(4) + 4(2) = 20$
Thus, the cartesian equation of $\Pi_1$ is $3y + 4z = 20$.

2. Find the cartesian equation of the second plane $\Pi_2$.

The normal vector for $\Pi_2$ is $\mathbf{n}_2 = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$.
The equation of $\Pi_2$ is of the form $x + 2y + 2z = c$.
Substitute the point $D(2,2,3)$ which lies on $\Pi_2$:
$c = (2) + 2(2) + 2(3) = 2 + 4 + 6 = 12$
Thus, the cartesian equation of $\Pi_2$ is $x + 2y + 2z = 12$.

3. Substitute the point $(p,0,q)$ into both plane equations to solve for $p$ and $q$.

Since $(p,0,q)$ lies in both planes, it must satisfy both cartesian equations.
Substitute $(p,0,q)$ into $\Pi_1$:
$3(0) + 4q = 20 \implies 4q = 20 \implies q = 5$

Substitute $(p,0,q)$ into $\Pi_2$:
$p + 2(0) + 2q = 12$
Since $q = 5$:
$p + 2(5) = 12 \implies p + 10 = 12 \implies p = 2$

1. Determine the direction vector of the line by finding the cross product of the normal vectors of the two planes.

From part (a), the normal to $\Pi_1$ is found to be $\mathbf{n}_1 = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix}$.
From the question stem, the normal to $\Pi_2$ is $\mathbf{n}_2 = \mathbf{i}+2\mathbf{j}+2\mathbf{k} = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$.

The direction vector of the line of intersection, $\mathbf{b}$, is perpendicular to both normal vectors:
$\mathbf{b} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 0 \\ 3 \\ 4 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} (3)(2) - (4)(2) \\ (4)(1) - (0)(2) \\ (0)(2) - (3)(1) \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \\ -3 \end{pmatrix}$

2. State the vector equation of the line using the common point found in part (c).

From part (c), the point $(2, 0, 5)$ lies on both planes and therefore lies on the line of intersection.
We can use its position vector as $\mathbf{a}$:
$\mathbf{a} = \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix}$

The equation of the line is:
$\mathbf{r} = \mathbf{a} + \lambda \mathbf{b}$
$\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ 4 \\ -3 \end{pmatrix}$

1. Find the direction vector of the line $AC$ and identify the normal vector of $\Pi_2$.

$\mathbf{d}_{AC} = \vec{OC} - \vec{OA} = \begin{pmatrix} 0 \\ 8 \\ -1 \end{pmatrix} - \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ -3 \end{pmatrix}$
The normal vector of $\Pi_2$ is $\mathbf{n}_2 = \begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix}$.

2. Set up the formula for the angle between a line and a plane.

Let $\theta$ be the acute angle between the line $AC$ and the plane $\Pi_2$.
$\sin \theta = \frac{|\mathbf{d}_{AC} \cdot \mathbf{n}_2|}{|\mathbf{d}_{AC}| |\mathbf{n}_2|}$
$\mathbf{d}_{AC} \cdot \mathbf{n}_2 = (-1)(1) + (4)(2) + (-3)(2) = -1 + 8 - 6 = 1$
$|\mathbf{d}_{AC}| = \sqrt{(-1)^2 + 4^2 + (-3)^2} = \sqrt{26}$
$|\mathbf{n}_2| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$
$\sin \theta = \frac{1}{3\sqrt{26}}$

3. Calculate the acute angle.

$\theta = \sin^{-1}\left( \frac{1}{3\sqrt{26}} \right) \approx 3.752^\circ$
Rounding to 1 decimal place, $\theta = 3.8^\circ$.