The planes $\pi_1$ and $\pi_2$, which meet in the line $l_1$, have equations
$\pi_1 : \mathbf{r} \cdot \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = 25 \text{ and } \pi_2 : x + ky - 2z = -15,$
where $k$ is a constant. Another line $l_2$ has equation $\mathbf{r} = \begin{pmatrix} -15 \\ 0 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}, \beta \in \mathbb{R}$.

1. Identify the condition for a point on $l_1$ to be independent of $k$.

The cartesian equations of the planes are:
$\pi_1: 2x - 3y + z = 25$
$\pi_2: x + ky - 2z = -15$
For the coordinates of a point $(x, y, z)$ on the intersection line $l_1$ to be independent of $k$, the term involving $k$ in the equation of $\pi_2$ must be zero. Hence, we must have $y = 0$.

2. Solve the resulting simultaneous equations to find the point's position vector.

Substituting $y = 0$ into the equations gives:
$2x + z = 25$
$x - 2z = -15$
From the second equation, $x = 2z - 15$. Substituting this into the first equation:
$2(2z - 15) + z = 25$
$5z - 30 = 25 \implies 5z = 55 \implies z = 11$
$x = 2(11) - 15 = 7$
The position vector of the point is $\begin{pmatrix} 7 \ 0 \ 11 \end{pmatrix}$.

3. Find the direction vector of $l_1$ using the cross product of the normal vectors.

The direction vector $\mathbf{d}$ of $l_1$ must be perpendicular to the normal vectors of both planes, $\mathbf{n}_1 = \begin{pmatrix} 2 \ -3 \ 1 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 1 \ k \ -2 \end{pmatrix}$.
$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 2 \ -3 \ 1 \end{pmatrix} \times \begin{pmatrix} 1 \ k \ -2 \end{pmatrix} = \begin{pmatrix} (-3)(-2) - (1)(k) \\ (1)(1) - (2)(-2) \\ (2)(k) - (-3)(1) \end{pmatrix} = \begin{pmatrix} 6 - k \\ 5 \\ 2k + 3 \end{pmatrix}$

4. Write the vector equation of the line $l_1$.

Using the position vector of the point and the direction vector, the vector equation of $l_1$ is:
$\mathbf{r} = \begin{pmatrix} 7 \ 0 \ 11 \end{pmatrix} + \lambda \begin{pmatrix} 6 - k \ 5 \ 2k + 3 \end{pmatrix}, \lambda \in \mathbb{R}$

1. Deduce the vector equation of line $l_1$ by finding a point on it and its direction vector.

From part (i), to find a point independent of $k$, we set $y=0$ in the equations of $\pi_1$ and $\pi_2$:
$2x + z = 25$
$x - 2z = -15$
Solving simultaneously yields $x=7$ and $z=11$, so the point $(7, 0, 11)$ lies on $l_1$.
The direction vector of $l_1$ is given by the cross product of the normal vectors of $\pi_1$ and $\pi_2$:
$\mathbf{d}_1 = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ k \\ -2 \end{pmatrix} = \begin{pmatrix} 6-k \\ 5 \\ 2k+3 \end{pmatrix}$
Thus, the vector equation of $l_1$ is:
$l_1: \mathbf{r} = \begin{pmatrix} 7 \\ 0 \\ 11 \end{pmatrix} + \alpha \begin{pmatrix} 6-k \\ 5 \\ 2k+3 \end{pmatrix}$

2. Set up a system of equations to test for the intersection of $l_1$ and $l_2$.

For the lines to intersect, there must exist scalars $\alpha$ and $\beta$ such that:
$\begin{pmatrix} 7 \\ 0 \\ 11 \end{pmatrix} + \alpha \begin{pmatrix} 6-k \\ 5 \\ 2k+3 \end{pmatrix} = \begin{pmatrix} -15 \\ 0 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$
Equating the corresponding components gives a system of three linear equations:
$7 + \alpha(6-k) = -15 + 3\beta \quad \text{--- (1)}$
$5\alpha = \beta \quad \text{--- (2)}$
$11 + \alpha(2k+3) = -\beta \quad \text{--- (3)}$

3. Solve the system of equations to determine the value of $k$ where the lines intersect.

Substitute (2) into (1):
$22 + \alpha(6-k) = 3(5\alpha) \implies 22 + 6\alpha - k\alpha = 15\alpha$
$\implies \alpha(k+9) = 22 \implies \alpha = \frac{22}{k+9}$
Substitute (2) into (3):
$11 + \alpha(2k+3) = -5\alpha \implies 11 + 2k\alpha + 3\alpha = -5\alpha$
$\implies \alpha(2k+8) = -11 \implies \alpha = -\frac{11}{2k+8}$
Equating the two expressions for $\alpha$:
$\frac{22}{k+9} = -\frac{11}{2k+8}$
Dividing by 11 and cross-multiplying:
$2(2k+8) = -(k+9)$
$4k + 16 = -k - 9$
$5k = -25 \implies k = -5$
The lines intersect exactly when $k = -5$.

4. Establish non-parallelism to conclude the exact values of $k$ for which the lines are skew.

Two lines are skew if they are neither parallel nor intersecting.
Checking for parallel lines:
$\begin{pmatrix} 6-k \\ 5 \\ 2k+3 \end{pmatrix} = c \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$
The $y$-components imply $c = 5$. For the $x$-components, $6-k = 15 \implies k = -9$. However, checking the $z$-components with $k = -9$ gives $2(-9)+3 = -15 \neq 5(-1)$, so the lines are never parallel for any real $k$.
Since $l_1$ and $l_2$ are never parallel, they are skew for all values of $k$ except where they intersect.
Therefore, the condition is $k \neq -5$.

1. Determine the direction vector and a point on $l_1$ by using the results from part (i).

From part (i), the point on $l_1$ independent of $k$ is found by setting $y=0$ in the planes' equations: $2x+z=25$ and $x-2z=-15$, giving $\mathbf{a}_1 = \begin{pmatrix} 7 \\ 0 \\ 11 \end{pmatrix}$.

The general direction vector is $\begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} \times \begin{pmatrix} 1 \\ k \\ -2 \end{pmatrix} = \begin{pmatrix} 6-k \\ 5 \\ 2k+3 \end{pmatrix}$.
Substituting $k=4$, the direction vector of $l_1$ is $\mathbf{d}_1 = \begin{pmatrix} 2 \\ 5 \\ 11 \end{pmatrix}$.
Thus, $l_1$ has the vector equation $\mathbf{r} = \begin{pmatrix} 7 \\ 0 \\ 11 \end{pmatrix} + \alpha \begin{pmatrix} 2 \\ 5 \\ 11 \end{pmatrix}$.

2. Identify $\overrightarrow{AB}$ as the shortest distance and find the common perpendicular direction.

Since $\overrightarrow{AB}$ is perpendicular to both $l_1$ and $l_2$, its length $|\overrightarrow{AB}|$ is the shortest distance between the two skew lines.
From $l_2$, a point is $\mathbf{a}_2 = \begin{pmatrix} -15 \\ 0 \\ 0 \end{pmatrix}$ and its direction vector is $\mathbf{d}_2 = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$.
The common perpendicular direction is given by the cross product of the direction vectors:
$\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{pmatrix} 2 \\ 5 \\ 11 \end{pmatrix} \times \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} -5-11 \\ 33-(-2) \\ 2-15 \end{pmatrix} = \begin{pmatrix} -16 \\ 35 \\ -13 \end{pmatrix}$

3. Calculate the length of $\overrightarrow{AB}$ using the projection formula.

The vector connecting the given points on the lines is $\mathbf{v} = \mathbf{a}_1 - \mathbf{a}_2 = \begin{pmatrix} 7 \\ 0 \\ 11 \end{pmatrix} - \begin{pmatrix} -15 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 22 \\ 0 \\ 11 \end{pmatrix}$.
The shortest distance is the magnitude of the projection of $\mathbf{v}$ onto $\mathbf{n}$:
$|\overrightarrow{AB}| = \frac{|\mathbf{v} \cdot \mathbf{n}|}{|\mathbf{n}|}$
$\mathbf{v} \cdot \mathbf{n} = \begin{pmatrix} 22 \\ 0 \\ 11 \end{pmatrix} \cdot \begin{pmatrix} -16 \\ 35 \\ -13 \end{pmatrix} = 22(-16) + 0(35) + 11(-13) = -495$
$|\mathbf{n}| = \sqrt{(-16)^2 + 35^2 + (-13)^2} = \sqrt{1650}$
$|\overrightarrow{AB}| = \frac{|-495|}{\sqrt{1650}} = \sqrt{ \frac{(-495)^2}{1650} } = \sqrt{ \frac{245025}{1650} } = \sqrt{ \frac{297}{2} }$
Thus, we have shown that $|\overrightarrow{AB}| = \sqrt{\frac{p}{2}}$, where $p = 297$.

1. Identify the direction vector of the line $l_2$ and the normal vector of the plane $\pi_1$.

The direction vector of $l_2$ is $\mathbf{d} = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}$.

The normal vector of $\pi_1$ is $\mathbf{n} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}$.

2. Calculate the dot product and the magnitudes of the two vectors.

$\mathbf{d} \cdot \mathbf{n} = (3)(2) + (1)(-3) + (-1)(1) = 6 - 3 - 1 = 2$

$|\mathbf{d}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{9 + 1 + 1} = \sqrt{11}$

$|\mathbf{n}| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$

3. Apply the geometric formula for the angle between a line and a plane.

Let $\theta$ be the acute angle between $l_2$ and $\pi_1$.

The sine of this angle is given by $\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}| |\mathbf{n}|}$.

$\sin \theta = \frac{|2|}{\sqrt{11}\sqrt{14}} = \frac{2}{\sqrt{154}}$

1. Establish the geometric relationship connecting the required distance to the previous parts.

Let the plane perpendicular to $l_1$ containing $l_2$ be $\pi_3$. The line $l_1$ intersects $\pi_3$ exactly at $A$ (from part iii), as $A$ lies on $l_1$ and $\overrightarrow{AB} \perp l_1, l_2$.

Since $P$ lies on $l_2 \subset \pi_3$ and $l_1 \perp \pi_3$ at $A$, the foot of the perpendicular from $P$ to $l_1$ is $A$. Hence, the required shortest distance from $P$ to $l_1$ is simply the length $AP$.

Moreover, since $l_1 \subset \pi_1$ and $l_1 \perp \pi_3$, it follows that $\pi_1 \perp \pi_3$. The line $AP$ represents the intersection of $\pi_1$ and $\pi_3$, meaning $AP$ is the perpendicular projection of $l_2$ onto $\pi_1$.
Therefore, the angle $\angle APB$ is exactly the acute angle $\theta$ between $l_2$ and $\pi_1$.

In the right-angled triangle $\triangle ABP$ (where $\angle ABP = 90^\circ$ because $AB \perp l_2$), we use the sine ratio:
$\sin \theta = \frac{AB}{AP}$

2. Calculate the length $AP$ using the deduced relationship and earlier answers.

From part (iii), the shortest distance between $l_1$ and $l_2$ is $AB = \sqrt{\frac{297}{2}}$.
From part (iv), the sine of the angle between $l_2$ and $\pi_1$ is $\sin \theta = \frac{2}{\sqrt{154}}$.

Substituting these values into the ratio:
$AP = \frac{AB}{\sin \theta} = \frac{\sqrt{\frac{297}{2}}}{\frac{2}{\sqrt{154}}}$
$AP = \frac{\sqrt{297} \sqrt{154}}{2\sqrt{2}}$
$AP = \frac{\sqrt{9 \times 33} \sqrt{77 \times 2}}{2\sqrt{2}}$
$AP = \frac{3\sqrt{33} \sqrt{77}}{2} = \frac{3\sqrt{3 \times 11 \times 7 \times 11}}{2}$
$AP = \frac{3 \times 11 \sqrt{21}}{2} = \frac{33\sqrt{21}}{2}$

The shortest distance from $P$ to $l_1$ is $\frac{33\sqrt{21}}{2}$.