the parametric equation is <-4,-2,2> + t<-6,-6,8> and the given plane equation is -8x-5y-2z=5

Thank you for the correction. Let's go through the problem again with the correct parametric equation of the line and the correct equation of the plane.

### Problem Restatement
We need to find the equation of a plane that:
1. Contains the line given by the parametric equation:
   \[
   \mathbf{r}(t) = \langle -4, -2, 2 \rangle + t \langle -6, -6, 8 \rangle
   \]
2. Is orthogonal to the plane given by the equation:
   \[
   -8x - 5y - 2z = 5
   \]

### Step 1: Identify the direction vector of the line
The direction vector of the line is \(\mathbf{v} = \langle -6, -6, 8 \rangle\).

### Step 2: Identify the normal vector of the given plane
The normal vector of the given plane \(-8x - 5y - 2z = 5\) is \(\mathbf{n}_1 = \langle -8, -5, -2 \rangle\).

### Step 3: Find the normal vector of the required plane
The plane we are looking for must be orthogonal to the given plane, so its normal vector \(\mathbf{n}_2\) must be perpendicular to \(\mathbf{n}_1\). Additionally, the plane must contain the line, meaning \(\mathbf{n}_2\) must also be perpendicular to the direction vector of the line \(\mathbf{v}\).

The normal vector \(\mathbf{n}_2\) can be found by taking the cross product of \(\mathbf{n}_1\) and \(\mathbf{v}\):
\[
\mathbf{n}_2 = \mathbf{n}_1 \times \mathbf{v}
\]
\[
\mathbf{n}_2 = \langle -8, -5, -2 \rangle \times \langle -6, -6, 8 \rangle
\]

### Step 4: Calculate the cross product
\[
\mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & -5 & -2 \\ -6 & -6 & 8 \end{vmatrix}
\]
\[
\mathbf{n}_2 = \mathbf{i} \left((-5)(8) - (-2)(-6)\right) - \mathbf{j} \left((-8)(8) - (-2)(-6)\right) + \mathbf{k} \left((-8)(-6) - (-5)(-6)\right)
\]
\[
\mathbf{n}_2 = \mathbf{i}(-40 - 12) - \mathbf{j}(-64 + 12) + \mathbf{k}(48 - 30)
\]
\[
\mathbf{n}_2 = \mathbf{i}(-52) - \mathbf{j}(-52) + \mathbf{k}(18)
\]
\[
\mathbf{n}_2 = \langle -52, 52, 18 \rangle
\]

### Step 5: Use a point on the line and the normal vector to find the equation of the plane
A point on the line is given by the position vector when \(t = 0\), which is \(\langle -4, -2, 2 \rangle\).

The equation of the plane can now be written as:
\[
-52(x + 4) + 52(y + 2) + 18(z - 2) = 0
\]

### Step 6: Simplify the equation
Expanding and simplifying:
\[
-52x - 208 + 52y + 104 + 18z - 36 = 0
\]
\[
-52x + 52y + 18z - 140 = 0
\]
We can divide the entire equation by \(-2\) to simplify it further:
\[
26x - 26y - 9z + 70 = 0
\]

This is the equation of the plane that satisfies the given conditions.

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

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Here are 5 related questions to explore further:

1. How do you determine if two vectors are perpendicular?
2. What is the geometric interpretation of a cross product between two vectors?
3. How can you find the distance from a point to a plane?
4. How do you find the line of intersection between two planes?
5. What conditions must be met for two planes to be parallel?

**Tip:** When simplifying the equation of a plane, dividing by the greatest common divisor of the coefficients can help keep the equation as simple as possible.

Learn how to find the equation of a plane that is orthogonal to a given line and contains another given plane. This problem involves parametric equations, vectors, and the equation of a plane. Suitable for undergraduate students studying linear algebra and vector calculus.

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