The lines $l_1$ and $l_2$ have equations $\mathbf{r} = \begin{pmatrix} 4 \\ 1 \\ 10 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ -1 \\ -6 \end{pmatrix}, \lambda \in \mathbb{R}$ and $\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 3 \\ 4 \end{pmatrix}, \mu \in \mathbb{R}$ respectively.

1. Identify the direction vectors of both lines

From the vector equations of the lines $l_1$ and $l_2$, their respective direction vectors are:
$\mathbf{d}_1 = \begin{pmatrix} 1 \\ -1 \\ -6 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} 4 \\ 3 \\ 4 \end{pmatrix}$.

2. Apply the scalar product formula to calculate the acute angle

Let $\theta$ be the acute angle between $l_1$ and $l_2$. Using the scalar product to find the angle between the direction vectors and taking the absolute value to ensure it is acute:
$\cos \theta = \frac{|\mathbf{d}_1 \cdot \mathbf{d}_2|}{|\mathbf{d}_1| |\mathbf{d}_2|}$
$\cos \theta = \frac{\left| \begin{pmatrix} 1 \\ -1 \\ -6 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 3 \\ 4 \end{pmatrix} \right|}{\sqrt{1^2 + (-1)^2 + (-6)^2} \sqrt{4^2 + 3^2 + 4^2}}$
$\cos \theta = \frac{|1(4) + (-1)(3) + (-6)(4)|}{\sqrt{38} \sqrt{41}}$
$\cos \theta = \frac{|4 - 3 - 24|}{\sqrt{1558}}$
$\cos \theta = \frac{|-23|}{\sqrt{1558}}$
$\theta = \cos^{-1}\left(\frac{23}{\sqrt{1558}}\right) \approx 54.359^\circ \approx 54.4^\circ \text{ (to 1 d.p.)}$

1. Calculate the direction vector of $l_3$ by finding the cross product of the direction vectors of $l_1$ and $l_2$.

Since $l_3$ is perpendicular to both $l_1$ and $l_2$, its direction vector $\mathbf{d_3}$ is parallel to the cross product of their direction vectors.
$\mathbf{d_3} = \begin{pmatrix} 1 \\ -1 \\ -6 \end{pmatrix} \times \begin{pmatrix} 4 \\ 3 \\ 4 \end{pmatrix}$
$= \begin{pmatrix} (-1)(4) - (-6)(3) \\ (-6)(4) - (1)(4) \\ (1)(3) - (-1)(4) \end{pmatrix}$
$= \begin{pmatrix} 14 \\ -28 \\ 7 \end{pmatrix}$
We can take a simpler parallel vector by dividing by 7:
$\mathbf{d} = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}$

2. Form the vector equation of line $l_3$.

$l_3$ passes through $A(1, -2, 4)$, so its position vector is $\begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}$.
Using a new parameter $t$, the vector equation is:
$\mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + t \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}, t \in \mathbb{R}$

1. Find a normal vector to the plane by taking the cross product of the direction vectors of $l_1$ and $l_2$ (or use the direction vector of $l_3$ from part ii).

The direction vectors of $l_1$ and $l_2$ are $\mathbf{d}_1 = \begin{pmatrix} 1 \\ -1 \\ -6 \end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix} 4 \\ 3 \\ 4 \end{pmatrix}$.
A normal vector to $p_1$ is $\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2$.
$\mathbf{n} = \begin{pmatrix} 1 \\ -1 \\ -6 \end{pmatrix} \times \begin{pmatrix} 4 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} -4 - (-18) \\ -24 - 4 \\ 3 - (-4) \end{pmatrix} = \begin{pmatrix} 14 \\ -28 \\ 7 \end{pmatrix}$
We can use a simpler, scaled normal vector: $\mathbf{n}' = \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}$.

2. Use a point on either line to find the scalar product form of the equation of the plane.

The plane contains $l_2$, so it passes through the point $(1, -2, 4)$.
The equation of the plane is $\mathbf{r} \cdot \mathbf{n}' = \mathbf{a} \cdot \mathbf{n}'$.
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}$
$\mathbf{r} \cdot \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} = 2(1) - 4(-2) + 1(4) = 2 + 8 + 4 = 14$

1. Form the vector equation of $l_3$ (incorporating results from part ii).

From part (ii), the direction vector of $l_3$ is perpendicular to both $l_1$ and $l_2$, found using the cross product of their direction vectors:
$\mathbf{d}_3 = \begin{pmatrix} 1 \\ -1 \\ -6 \end{pmatrix} \times \begin{pmatrix} 4 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 14 \\ -28 \\ 7 \end{pmatrix} = 7 \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}$
Since $l_3$ passes through $A(1, -2, 4)$, its vector equation is:
$l_3: \mathbf{r} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + t\begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix}, t \in \mathbb{R}$

2. Show that the point $B$ lies on $l_3$.

Equating the position vector of $B(-5, 10, 1)$ to the equation of $l_3$:
$\begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} + t\begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} = \begin{pmatrix} -5 \\ 10 \\ 1 \end{pmatrix}$
Comparing the $x$-components:
$1 + 2t = -5 \implies 2t = -6 \implies t = -3$
Substitute $t = -3$ into the $y$ and $z$ components to check for consistency:
$y$-component: $-2 - 4(-3) = -2 + 12 = 10$
$z$-component: $4 + 1(-3) = 4 - 3 = 1$
Since a single value of $t = -3$ yields all three correct coordinates, point $B$ lies on $l_3$.

3. Deduce the position vector of the point of intersection of $l_3$ and $p_1$.

The plane $p_1$ is defined to contain line $l_2$, and point $A(1, -2, 4)$ is the position vector given in the equation of $l_2$, so $A$ lies on $p_1$.
Since $l_3$ is explicitly constructed to pass through $A$, $A$ is a point common to both $l_3$ and $p_1$.
Thus, the position vector of the point of intersection of $l_3$ and $p_1$ is $\begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}$.

4. Calculate the exact perpendicular distance from $B$ to $p_1$.

Since $l_3$ is perpendicular to both $l_1$ and $l_2$, it is parallel to the normal of plane $p_1$. Hence, $l_3$ is the perpendicular line from $B$ to $p_1$.
The perpendicular distance from $B$ (which is on $l_3$) to $p_1$ is simply the distance between $B$ and the point of intersection $A$:
$|\vec{AB}| = \left| \begin{pmatrix} 1 - (-5) \\ -2 - 10 \\ 4 - 1 \end{pmatrix} \right| = \left| \begin{pmatrix} 6 \\ -12 \\ 3 \end{pmatrix} \right|$
$|\vec{AB}| = \sqrt{6^2 + (-12)^2 + 3^2} = \sqrt{36 + 144 + 9} = \sqrt{189} = \sqrt{9 \times 21} = 3\sqrt{21}$
The exact perpendicular distance is $3\sqrt{21}$.