We are tasked with finding a plane that contains the line $\vec{r}(t) = \langle -6, -7, 3 \rangle + t \langle -8, 4, -2 \rangle$ and is orthogonal to the plane given by the equation:

$-1x + 8y - 5z = -4$

Step-by-Step Solution:

1. Understand the Line Equation:
   The parametric equation of the line is given as: $\vec{r}(t) = \langle -6, -7, 3 \rangle + t \langle -8, 4, -2 \rangle$ Here, $\langle -6, -7, 3 \rangle$ is a point on the line, and $\langle -8, 4, -2 \rangle$ is the direction vector of the line.

2. Normal Vector of the Plane:
   The plane is orthogonal to the plane $-x + 8y - 5z = -4$. The normal vector to this plane is the coefficient vector of $x$, $y$, and $z$, which is: $\vec{n}_1 = \langle -1, 8, -5 \rangle$

3. Find the Normal Vector of the Desired Plane:
   The plane we are looking for contains the line, so it must have a normal vector orthogonal to the direction vector of the line. Let the normal vector of the desired plane be $\vec{n}_2$.

   The plane is also orthogonal to the given plane $-x + 8y - 5z = -4$, which implies that $\vec{n}_2$ must be perpendicular to the normal vector of the given plane, $\vec{n}_1$. Therefore, the normal vector of the desired plane is the cross product of the direction vector of the line and the normal vector of the given plane.

4. Calculate the Cross Product:
   Let $\vec{d} = \langle -8, 4, -2 \rangle$ be the direction vector of the line. Now, we compute the cross product: $\vec{n}_2 = \vec{d} \times \vec{n}_1$

   Let's compute the cross product step by step.

I'll calculate this next.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, you can calculate the cross product manually using the following formula:

$\vec{n}_2 = \vec{d} \times \vec{n}_1$ where $\vec{d} = \langle -8, 4, -2 \rangle$ and $\vec{n}_1 = \langle -1, 8, -5 \rangle$.

The formula for cross product is: $\vec{d} \times \vec{n}_1 = \left( \begin{vmatrix} 4 & -2 \\ 8 & -5 \end{vmatrix}, -\begin{vmatrix} -8 & -2 \\ -1 & -5 \end{vmatrix}, \begin{vmatrix} -8 & 4 \\ -1 & 8 \end{vmatrix} \right)$ This will give you the normal vector of the desired plane.

Would you like assistance with the calculation?

Here are 5 follow-up questions to enhance your understanding:

1. How is the normal vector of a plane related to its equation?
2. Can the same cross-product approach be used to find other orthogonal planes?
3. What are the geometric implications of two planes being orthogonal?
4. How do you derive a plane equation from a normal vector and a point?
5. Why is the cross product important for finding perpendicular vectors in 3D space?

Tip: When dealing with multiple planes and lines, always check how their vectors interact geometrically (dot and cross products).