Given the line with parametric equatio x = 3+t, y=5+2t,z=7-tFind the equations of the two planes which intersection is the given line.

Given the line with parametric equations:

\[
x = 3 + t, \quad y = 5 + 2t, \quad z = 7 - t
\]

we need to find the equations of two planes whose intersection is this line.

### Step 1: Find the direction vector of the line

The direction vector \(\mathbf{v}\) of the line can be determined from the coefficients of \(t\) in the parametric equations:

\[
\mathbf{v} = \langle 1, 2, -1 \rangle
\]

### Step 2: Choose a point on the line

Let's choose a specific value of \(t\) to find a point on the line. If \(t = 0\), the point \(\mathbf{P}\) on the line is:

\[
\mathbf{P}(3, 5, 7)
\]

### Step 3: Define the two planes

We want two planes whose intersection is the line defined by the given parametric equations. We know the direction vector \(\mathbf{v} = \langle 1, 2, -1 \rangle\) should be perpendicular to the normal vectors of these planes.

Let's assume the normal vectors of the two planes are \(\mathbf{n_1}\) and \(\mathbf{n_2}\). We require:

\[
\mathbf{n_1} \cdot \mathbf{v} = 0 \quad \text{and} \quad \mathbf{n_2} \cdot \mathbf{v} = 0
\]

Let's choose two normal vectors that satisfy this condition. For simplicity, we can select:

\[
\mathbf{n_1} = \langle 2, -1, 0 \rangle \quad \text{and} \quad \mathbf{n_2} = \langle 1, 0, 1 \rangle
\]

### Step 4: Write the equations of the planes

The general equation of a plane is:

\[
ax + by + cz = d
\]

For Plane 1 with normal vector \(\mathbf{n_1} = \langle 2, -1, 0 \rangle\):

\[
2(x - 3) - 1(y - 5) + 0(z - 7) = 0
\]

Simplifying:

\[
2x - y = 1 \quad \text{(Equation of Plane 1)}
\]

For Plane 2 with normal vector \(\mathbf{n_2} = \langle 1, 0, 1 \rangle\):

\[
1(x - 3) + 0(y - 5) + 1(z - 7) = 0
\]

Simplifying:

\[
x + z = 10 \quad \text{(Equation of Plane 2)}
\]

### Final Answer

The equations of the two planes whose intersection is the given line are:

1. \(2x - y = 1\)
2. \(x + z = 10\)

Would you like any more details or have any questions about this solution?

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Here are some related questions:

1. How do you find the normal vector of a plane?
2. What is the geometric interpretation of the direction vector of a line?
3. How can you find the intersection of two planes?
4. How do you convert a parametric equation of a line into its vector form?
5. What is the relationship between the normal vector and the plane's equation?
6. How do you determine if two lines are parallel, intersecting, or skew?
7. How can you find the distance between a point and a plane?
8. What are the steps to find the intersection line between two planes?

**Tip:** When working with parametric equations, always identify the direction vector as it provides critical information about the orientation of the line.

Learn how to find equations of planes intersecting a given line with parametric equations x = 3 + t, y = 5 + 2t, z = 7 - t. Step-by-step solution involving direction vectors, normal vectors, and plane equations.

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