To determine the relationship between the two lines (whether they are parallel, intersect, or are skew), we can follow these steps:

1. Find Direction Vectors for each line.
2. Check for Parallelism by comparing the direction vectors.
3. If they are not parallel, check for intersection by setting up parametric equations for both lines and solving for any common point.

Step 1: Find Direction Vectors

For each line, the direction vector can be determined by subtracting the coordinates of one point from the other.

Line 1

Given points: $(0, 5, 0)$ and $(-1, 4, 3)$.

• Direction vector $\mathbf{d_1} = (-1 - 0, 4 - 5, 3 - 0) = (-1, -1, 3)$.

Line 2

Given points: $(-4, 2, 5)$ and $(-1, 11, 9)$.

• Direction vector $\mathbf{d_2} = (-1 - (-4), 11 - 2, 9 - 5) = (3, 9, 4)$.

Step 2: Check for Parallelism

To check if the lines are parallel, we need to see if one direction vector is a scalar multiple of the other. This would mean there exists a constant $k$ such that: $\mathbf{d_1} = k \cdot \mathbf{d_2}$ Comparing $(-1, -1, 3)$ and $(3, 9, 4)$, it is evident that there is no scalar $k$ that satisfies all three components. Therefore, the lines are not parallel.

Step 3: Check for Intersection

Since the lines are not parallel, we’ll determine if they intersect by setting up parametric equations and solving for any common point.

Parametric Equations

For Line 1 passing through point $(0, 5, 0)$ with direction vector $(-1, -1, 3)$, we can write: $(x_1, y_1, z_1) = (0, 5, 0) + t(-1, -1, 3) = (-t, 5 - t, 3t)$

For Line 2 passing through point $(-4, 2, 5)$ with direction vector $(3, 9, 4)$, we can write: $(x_2, y_2, z_2) = (-4, 2, 5) + s(3, 9, 4) = (-4 + 3s, 2 + 9s, 5 + 4s)$

Set up Equations for Intersection

For the lines to intersect, we must have:

1. $-t = -4 + 3s$
2. $5 - t = 2 + 9s$
3. $3t = 5 + 4s$

Solving these equations simultaneously:

1. From the first equation: $t = 4 - 3s$.
2. Substitute $t = 4 - 3s$ into the second equation: $5 - (4 - 3s) = 2 + 9s$ $1 + 3s = 2 + 9s$ $-1 = 6s \Rightarrow s = -\frac{1}{6}$
3. Substitute $s = -\frac{1}{6}$ back into $t = 4 - 3s$: $t = 4 - 3\left(-\frac{1}{6}\right) = 4 + \frac{1}{2} = \frac{9}{2}$

Verify Intersection Point

Using $t = \frac{9}{2}$ in Line 1: [ (x_1, y_1, z_1) = \left(-\frac{9}{2}, 5 - \frac{9}{2}, 3 \cdot \frac{9}{2}\right) = \left(-\frac{9}{2}, -\frac{-1}{2}, \frac{27}{2}\right)