The planes $\Pi_1$ and $\Pi_2$ have equations $\mathbf{r} \cdot (\mathbf{i} - 2\mathbf{j} + \mathbf{k}) = 4$ and $\mathbf{r} = 3\mathbf{i} - \mathbf{j} + \mathbf{k} + \lambda(-\mathbf{i} + \mathbf{j} + \mathbf{k}) + \mu(4\mathbf{i} + \mathbf{j})$ respectively, and meet in a line $l_1$.

1. Find the normal vectors for both planes.

For $\Pi_1$, the normal vector is $\mathbf{n}_1 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$.
For $\Pi_2$, the normal vector $\mathbf{n}_2$ is the cross product of its two direction vectors:
$\mathbf{n}_2 = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} (1)(0) - (1)(1) \\ (1)(4) - (-1)(0) \\ (-1)(1) - (1)(4) \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ -5 \end{pmatrix}$

2. Calculate the scalar product and magnitudes of the normal vectors.

$\mathbf{n}_1 \cdot \mathbf{n}_2 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 4 \\ -5 \end{pmatrix} = (1)(-1) + (-2)(4) + (1)(-5) = -1 - 8 - 5 = -14$
$|\mathbf{n}_1| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$
$|\mathbf{n}_2| = \sqrt{(-1)^2 + 4^2 + (-5)^2} = \sqrt{1 + 16 + 25} = \sqrt{42}$

3. Find the acute angle between the planes using the scalar product formula.

Let $\theta$ be the acute angle between the planes.
$\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$
$\cos \theta = \frac{|-14|}{\sqrt{6}\sqrt{42}} = \frac{14}{\sqrt{252}} = \frac{14}{6\sqrt{7}} = \frac{7}{3\sqrt{7}} = \frac{\sqrt{7}}{3}$
$\theta = \cos^{-1}\left(\frac{\sqrt{7}}{3}\right) \approx 28.1255^\circ$
$\theta \approx 28.1^\circ \text{ (to 1 d.p.)}$

1. Substitute the parametric equation of $\Pi_2$ into the scalar product equation of $\Pi_1$ to find the relationship between $\lambda$ and $\mu$.

The equation of $\Pi_1$ is $\mathbf{r} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = 4$.

Substitute $\mathbf{r}$ from $\Pi_2$:
$\left[ \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix} \right] \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} = 4$

Evaluating the dot products:
$(3(1) - 1(-2) + 1(1)) + \lambda(-1(1) + 1(-2) + 1(1)) + \mu(4(1) + 1(-2) + 0(1)) = 4$
$(3 + 2 + 1) + \lambda(-1 - 2 + 1) + \mu(4 - 2 + 0) = 4$
$6 - 2\lambda + 2\mu = 4$
$-2\lambda + 2\mu = -2$
$\lambda = \mu + 1$

2. Substitute the relationship back into the equation of $\Pi_2$ to obtain the vector equation of the line of intersection $l_1$.

Substitute $\lambda = \mu + 1$ into the equation of $\Pi_2$:
$\mathbf{r} = \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix} + (\mu + 1) \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix}$

Group the constant vectors and the vectors with parameter $\mu$:
$\mathbf{r} = \left[ \begin{pmatrix} 3 \\ -1 \\ 1 \end{pmatrix} + \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \right] + \mu \left[ \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} + \begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix} \right]$
$\mathbf{r} = \begin{pmatrix} 2 \\ 0 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}, \quad \mu \in \mathbb{R}$

1. Find the normal to plane $\Pi_2$ and the direction vector of the line of intersection $l_1$ (required from part ii).

From part (ii), the normal to $\Pi_1$ is $\mathbf{n}_1 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$.
The normal to $\Pi_2$ is found by the cross product of its direction vectors:
$\mathbf{n}_2 = \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 4 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} -1 \\ 4 \\ -5 \end{pmatrix}$
The direction vector of the line $l_1$ is parallel to $\mathbf{n}_1 \times \mathbf{n}_2$:
$\mathbf{d} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \times \begin{pmatrix} -1 \\ 4 \\ -5 \end{pmatrix} = \begin{pmatrix} 6 \\ 4 \\ 2 \end{pmatrix} = 2\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$
We can use the simplified direction vector $\mathbf{m} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$.

2. Find the vector $\overrightarrow{AB}$.

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} - \begin{pmatrix} 6 \\ 3 \\ -5 \end{pmatrix} = \begin{pmatrix} -4 \\ 0 \\ 6 \end{pmatrix}$

3. Calculate the length of projection of $\overrightarrow{AB}$ onto $l_1$.

The length of projection of $\overrightarrow{AB}$ onto $l_1$ is given by $\frac{|\overrightarrow{AB} \cdot \mathbf{m}|}{|\mathbf{m}|}$:
$\text{Length} = \frac{\left| \begin{pmatrix} -4 \\ 0 \\ 6 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \right|}{\sqrt{3^2 + 2^2 + 1^2}}$
$\text{Length} = \frac{|(-4)(3) + (0)(2) + (6)(1)|}{\sqrt{14}} = \frac{|-12 + 6|}{\sqrt{14}}$
$\text{Length} = \frac{6}{\sqrt{14}} = \frac{6\sqrt{14}}{14} = \frac{3\sqrt{14}}{7}$

1. Set up an equation for $p$ using the perpendicular distance from point $C$ to plane $\Pi_1$.

The position vector of $C$ is $\mathbf{c} = \begin{pmatrix} p \\ 2p + 1 \\ -3 \end{pmatrix}$ and the normal vector to $\Pi_1$ is $\mathbf{n}_1 = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$.

The perpendicular distance from $C$ to $\Pi_1$ is given by $\frac{|\mathbf{c} \cdot \mathbf{n}_1 - D|}{|\mathbf{n}_1|}$.

$\frac{\left| \begin{pmatrix} p \\ 2p + 1 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} - 4 \right|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{15}{\sqrt{6}}$

$\frac{|p - 2(2p + 1) - 3 - 4|}{\sqrt{6}} = \frac{15}{\sqrt{6}}$

$|-3p - 9| = 15$

2. Solve for $p$, noting that $p$ is a positive constant.

$|-3(p + 3)| = 15$

$3|p + 3| = 15$

$|p + 3| = 5$

$p + 3 = 5 \implies p = 2$

OR

$p + 3 = -5 \implies p = -8$

Since $p$ is a positive constant, $p = 2$.

3. Set up an equation for $q$ using the acute angle between line $l_2$ and plane $\Pi_1$.

The direction vector of $l_2$ is $\mathbf{d}_2 = \begin{pmatrix} 3q \\ -3 \\ q \end{pmatrix}$. Let $\theta$ be the acute angle between $l_2$ and $\Pi_1$.

$\sin \theta = \frac{|\mathbf{d}_2 \cdot \mathbf{n}_1|}{|\mathbf{d}_2| |\mathbf{n}_1|}$

Given $\theta = \sin^{-1}\left(\frac{2}{\sqrt{6}}\right)$, we have $\sin \theta = \frac{2}{\sqrt{6}}$.

$\frac{\left| \begin{pmatrix} 3q \\ -3 \\ q \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} \right|}{\sqrt{(3q)^2 + (-3)^2 + q^2} \sqrt{6}} = \frac{2}{\sqrt{6}}$

$\frac{|3q + 6 + q|}{\sqrt{10q^2 + 9} \sqrt{6}} = \frac{2}{\sqrt{6}}$

$|4q + 6| = 2\sqrt{10q^2 + 9}$

4. Solve for $q$, noting that $q$ is a positive constant.

Divide by 2 on both sides:

$|2q + 3| = \sqrt{10q^2 + 9}$

Square both sides:

$(2q + 3)^2 = 10q^2 + 9$

$4q^2 + 12q + 9 = 10q^2 + 9$

$6q^2 - 12q = 0$

$6q(q - 2) = 0$

$q = 0$ or $q = 2$.

Since $q$ is a positive constant, $q = 2$.