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The image appears to contain a mathematical problem that involves finding parametric equations of a line passing through a given point and perpendicular to a given plane. Additionally, the task includes finding the distance between a point and the plane.

To solve the problem, I'll first extract the text from the image:

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**Problem Statement:**
1. **Find parametric equations of the line that contains the point P(2, -1, 4) and is perpendicular to the plane 6x - 2y + z = 2.**
2. **Find the distance between P and the plane.**

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### Step-by-Step Solution

#### 1. Finding the Parametric Equations of the Line

A line that passes through the point \( P(2, -1, 4) \) and is perpendicular to the plane \( 6x - 2y + z = 2 \) will have a direction vector that is normal to the plane. The normal vector \( \mathbf{n} \) of the plane can be obtained from the coefficients of \( x, y, \) and \( z \) in the plane equation:

\[
\mathbf{n} = (6, -2, 1)
\]

The parametric equations of a line passing through point \( P(x_0, y_0, z_0) = (2, -1, 4) \) with direction vector \( \mathbf{d} = (a, b, c) \) are:

\[
x = x_0 + at, \; y = y_0 + bt, \; z = z_0 + ct
\]

Substituting the values:

\[
x = 2 + 6t, \; y = -1 - 2t, \; z = 4 + t
\]

#### 2. Finding the Distance Between Point P and the Plane

The formula to find the distance \( D \) from a point \( (x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \) is:

\[
D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]

Given the plane equation \( 6x - 2y + z = 2 \), we have \( A = 6 \), \( B = -2 \), \( C = 1 \), and \( D = -2 \).

Substitute point \( P(2, -1, 4) \):

\[
D = \frac{|6(2) - 2(-1) + 1(4) - 2|}{\sqrt{6^2 + (-2)^2 + 1^2}}
\]

Calculating the numerator:

\[
D = \frac{|12 + 2 + 4 - 2|}{\sqrt{36 + 4 + 1}} = \frac{|16|}{\sqrt{41}}
\]

Thus:

\[
D = \frac{16}{\sqrt{41}}
\]

To obtain the decimal form, compute:

\[
D \approx \frac{16}{6.4} \approx 2.5
\]

### Final Answers:

1. **Parametric Equations of the Line:**
   \[
   x = 2 + 6t, \; y = -1 - 2t, \; z = 4 + t
   \]

2. **Distance Between Point P and the Plane:**
   \[
   D = \frac{16}{\sqrt{41}} \approx 2.5
   \]

Would you like more details, or have any further questions?

### Additional Questions to Explore:

1. How do you find the normal vector to a plane from its equation?
2. Can you explain the general formula for the distance from a point to a plane?
3. What are some other methods to find a line's parametric equations?
4. How can you verify if two lines are perpendicular or parallel in 3D space?
5. How does the concept of vectors relate to lines and planes in 3D geometry?

**Tip:** To handle similar problems efficiently, remember that the coefficients of a plane equation always represent a normal vector to that plane.

Find parametric equations of the line that contains the point P(2, -1, 4) and is perpendicular to the plane 6x - 2y + z = 2. Also, find the distance between P and the plane.

Learn how to find the parametric equations of a line passing through point P(2, -1, 4) and perpendicular to the plane 6x - 2y + z = 2. Additionally, discover how to calculate the distance between point P and the plane using key formulas. Suitable for students in Grades 10-12.

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