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.