Determine if the following lines are parallel, intersecting, or skew: L: (x+8)/2 = (y−10)/3 = (z−6)/1 and M: (x-1)/-1 = (y−1)/2 = (z−1)/4

For line L, the direction vector is $\vec{d_L} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$ and a point on the line is $P_L = (-8, 10, 6)$.
For line M, the direction vector is $\vec{d_M} = \begin{pmatrix} -1 \\ 2 \\ 4 \end{pmatrix}$ and a point on the line is $P_M = (1, 1, 1)$.

Lines are parallel if their direction vectors are scalar multiples of each other.
We check if $\vec{d_L} = k \vec{d_M}$ for some scalar $k$.

From the x-components: $2 = -k \implies k = -2$.
From the y-components: $3 = 2k \implies k = \frac{3}{2}$.
Since the values of $k$ are not consistent ($-2 \neq \frac{3}{2}$), the direction vectors are not parallel. Therefore, the lines are not parallel.

To check for intersection, we write the parametric equations for each line and set the corresponding coordinates equal.
Parametric equations for L:
$x_L = -8 + 2t$
$y_L = 10 + 3t$
$z_L = 6 + t$

Parametric equations for M:
$x_M = 1 - s$
$y_M = 1 + 2s$
$z_M = 1 + 4s$

Set the coordinates equal:

1. $-8 + 2t = 1 - s$
2. $10 + 3t = 1 + 2s$
3. $6 + t = 1 + 4s$

From equation (3), express $t$ in terms of $s$:
$t = 1 + 4s - 6$
$t = 4s - 5$

Substitute this expression for $t$ into equation (1):
$-8 + 2(4s - 5) = 1 - s$
$-8 + 8s - 10 = 1 - s$
$8s - 18 = 1 - s$
$9s = 19$
$s = \frac{19}{9}$

Now substitute $s = \frac{19}{9}$ back into the expression for $t$:
$t = 4\left(\frac{19}{9}\right) - 5$
$t = \frac{76}{9} - \frac{45}{9}$
$t = \frac{31}{9}$

Finally, check if these values of $s$ and $t$ satisfy equation (2):
$10 + 3t = 1 + 2s$
$10 + 3\left(\frac{31}{9}\right) = 1 + 2\left(\frac{19}{9}\right)$
$10 + \frac{31}{3} = 1 + \frac{38}{9}$
$\frac{30}{3} + \frac{31}{3} = \frac{9}{9} + \frac{38}{9}$
$\frac{61}{3} = \frac{47}{9}$
This statement is false, as $\frac{61}{3} = \frac{183}{9}$. Since the values of $s$ and $t$ do not satisfy all three equations, the lines do not intersect.

Since the lines are not parallel and do not intersect, they are skew.
We can also check if they are perpendicular by calculating the dot product of their direction vectors:
$\vec{d_L} \cdot \vec{d_M} = (2)(-1) + (3)(2) + (1)(4) = -2 + 6 + 4 = 8$.
Since the dot product is not zero, the lines are not perpendicular.

The relationship between lines L and M is that they are skew lines.

The final answer is $\text{Skew lines}$