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Let's solve the given problem. We need to determine the relationship of the line \( L(t) = \langle -1 - 5t, -1 - 4t, 4 - 3t \rangle \) with each of the given planes.

### Step 1: Extract Direction Vector of the Line
The direction vector of the line \( L(t) \) is obtained from the coefficients of \( t \):
\[
\vec{d} = \langle -5, -4, -3 \rangle
\]

### Step 2: Find Normal Vectors of the Planes
Each plane equation is given in the form \( ax + by + cz = d \). The normal vector of a plane is given by \( \vec{n} = \langle a, b, c \rangle \).

#### Plane 1: \( 7.5x + 6y + 4.5z = -67.5 \)
- Normal vector: \(\vec{n_1} = \langle 7.5, 6, 4.5 \rangle\)

#### Plane 2: \( 10x + 8y + 6z = -10 \)
- Normal vector: \(\vec{n_2} = \langle 10, 8, 6 \rangle\)

#### Plane 3: \( 14x - 16y - 2z = -10 \)
- Normal vector: \(\vec{n_3} = \langle 14, -16, -2 \rangle\)

#### Plane 4: \( x - 3y + 4z = 13 \)
- Normal vector: \(\vec{n_4} = \langle 1, -3, 4 \rangle\)

### Step 3: Determine Relationship Between Line and Each Plane
To determine if the line is parallel, perpendicular, or neither with respect to a plane, we use the direction vector of the line and the normal vector of the plane:

1. **Perpendicular Line to Plane**: The line is perpendicular to the plane if the dot product of the direction vector and the normal vector is zero:
   \[
   \vec{d} \cdot \vec{n} = 0
   \]

2. **Parallel Line to Plane**: The line is parallel to the plane if the direction vector is orthogonal to the normal vector of the plane, which means the dot product is zero.

3. **Neither**: If the line is neither perpendicular nor parallel, it falls into the "neither" category.

#### Calculations
##### Plane 1
\[
\vec{d} \cdot \vec{n_1} = (-5)(7.5) + (-4)(6) + (-3)(4.5) = -37.5 - 24 - 13.5 = -75
\]
Since \(\vec{d} \cdot \vec{n_1} \neq 0\), the line is **neither parallel nor perpendicular** to Plane 1.

##### Plane 2
\[
\vec{d} \cdot \vec{n_2} = (-5)(10) + (-4)(8) + (-3)(6) = -50 - 32 - 18 = -100
\]
Since \(\vec{d} \cdot \vec{n_2} \neq 0\), the line is **neither parallel nor perpendicular** to Plane 2.

##### Plane 3
\[
\vec{d} \cdot \vec{n_3} = (-5)(14) + (-4)(-16) + (-3)(-2) = -70 + 64 + 6 = 0
\]
Since \(\vec{d} \cdot \vec{n_3} = 0\), the line is **parallel** to Plane 3.

##### Plane 4
\[
\vec{d} \cdot \vec{n_4} = (-5)(1) + (-4)(-3) + (-3)(4) = -5 + 12 - 12 = -5
\]
Since \(\vec{d} \cdot \vec{n_4} \neq 0\), the line is **neither parallel nor perpendicular** to Plane 4.

### Summary of Relationships
- **Plane 1** (\( 7.5x + 6y + 4.5z = -67.5 \)): **Neither**
- **Plane 2** (\( 10x + 8y + 6z = -10 \)): **Neither**
- **Plane 3** (\( 14x - 16y - 2z = -10 \)): **Parallel**
- **Plane 4** (\( x - 3y + 4z = 13 \)): **Neither**

Do you have any questions or would you like further details?

### Related Questions:
1. How can we determine if a line lies entirely within a plane?
2. What is the significance of the cross product in determining relationships between lines and planes?
3. How would you find the intersection point of a line and a plane?
4. Could you explain how to parameterize the equation of a line given two points?
5. How do the direction vector and the normal vector relate in the context of skew lines?

**Tip:** When determining if two vectors are perpendicular, use the dot product. If the result is zero, they are perpendicular!

Consider the line L(t) = ⟨−1−5t,−1−4t,4−3t⟩. Then, determine if the line is parallel, perpendicular, or neither with respect to the planes given by the equations: 7.5x + 6y + 4.5z = -67.5, 10x + 8y + 6z = -10, 14x - 16y - 2z = -10, and x - 3y + 4z = 13.

Learn how to determine whether the line L(t) = ⟨−1−5t,−1−4t,4−3t⟩ is parallel, perpendicular, or neither with respect to the planes 7.5x + 6y + 4.5z = -67.5, 10x + 8y + 6z = -10, 14x - 16y - 2z = -10, and x - 3y + 4z = 13. Explore vector algebra and the use of the dot product in 3D geometry.

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