The lines l_1 and l_2 have equations $\mathbf{r}=\begin{pmatrix}2.5k\\k\\2k\end{pmatrix}+\lambda\begin{pmatrix}1\\3\\1\end{pmatrix}\quad\text{and}\quad \mathbf{r}=\begin{pmatrix}3\\0.5k\\0.5k\end{pmatrix}+\mu\begin{pmatrix}4\\7\\5\end{pmatrix},\;\lambda,\mu\in\mathbb{R}$ respectively, where k is a constant.

1. Set up a system of linear equations by equating the components of the position vectors for $l_1$ and $l_2$.

Since the lines intersect, there must exist scalars $\lambda$ and $\mu$ such that:
$\begin{pmatrix} 2.5k \\ k \\ 2k \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 0.5k \\ 0.5k \end{pmatrix} + \mu \begin{pmatrix} 4 \\ 7 \\ 5 \end{pmatrix}$

Equating the $x$, $y$, and $z$ components gives three equations:

1. $2.5k + \lambda = 3 + 4\mu \implies \lambda - 4\mu = 3 - 2.5k$
2. $k + 3\lambda = 0.5k + 7\mu \implies 3\lambda - 7\mu = -0.5k$
3. $2k + \lambda = 0.5k + 5\mu \implies \lambda - 5\mu = -1.5k$

2. Solve two of the equations to express $\lambda$ and $\mu$ in terms of $k$.

Subtracting equation (3) from equation (1):
$(\lambda - 4\mu) - (\lambda - 5\mu) = (3 - 2.5k) - (-1.5k)$
$\mu = 3 - k$

Substitute $\mu = 3 - k$ back into equation (1):
$\lambda - 4(3 - k) = 3 - 2.5k$
$\lambda - 12 + 4k = 3 - 2.5k$
$\lambda = 15 - 6.5k$

3. Substitute the expressions for $\lambda$ and $\mu$ into the remaining equation to solve for $k$.

Substitute $\lambda$ and $\mu$ into equation (2):
$3(15 - 6.5k) - 7(3 - k) = -0.5k$
$45 - 19.5k - 21 + 7k = -0.5k$
$24 - 12.5k = -0.5k$
$12k = 24$
$k = 2$

1. Determine the values of $k$ and the parameter $\lambda$ or $\mu$ from the intersection of the two lines (as required from part (a)).

Equating the components of $\mathbf{r}$ for $l_1$ and $l_2$:
$2.5k + \lambda = 3 + 4\mu \quad \text{--- (1)}$
$k + 3\lambda = 0.5k + 7\mu \quad \text{--- (2)}$
$2k + \lambda = 0.5k + 5\mu \quad \text{--- (3)}$
Subtracting (3) from (1) gives $\mu = 3 - k$.
Substituting $\mu = 3 - k$ into (3) yields $\lambda = 15 - 6.5k$.
Substituting both expressions into (2) gives:
$k + 3(15 - 6.5k) = 0.5k + 7(3 - k)$
$45 - 18.5k = 21 - 6.5k$
$12k = 24 \implies k = 2$
Using $k = 2$, we find $\mu = 3 - 2 = 1$.

2. Substitute $k$ and $\mu$ into the equation of $l_2$ to find the point of intersection.

Using the equation for $l_2$ with $k = 2$ and $\mu = 1$:
$\mathbf{r} = \begin{pmatrix} 3 \\ 0.5(2) \\ 0.5(2) \end{pmatrix} + 1 \begin{pmatrix} 4 \\ 7 \\ 5 \end{pmatrix}$
$\mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 1 \end{pmatrix} + \begin{pmatrix} 4 \\ 7 \\ 5 \end{pmatrix} = \begin{pmatrix} 7 \\ 8 \\ 6 \end{pmatrix}$
The coordinates of the point of intersection are $(7, 8, 6)$.