To find the equation of a plane that passes through a given point $P = (5, 5, 5)$ and is parallel to a given plane $5x + 5y + 4z = -6$, we can use the fact that parallel planes have the same normal vector.

The given plane $5x + 5y + 4z = -6$ has a normal vector $\mathbf{n} = \langle 5, 5, 4 \rangle$.

Since the new plane is parallel to the given plane, it will have the same normal vector. The general equation of a plane with normal vector $\langle A, B, C \rangle$ is:

$Ax + By + Cz = D$

Using the normal vector $\mathbf{n} = \langle 5, 5, 4 \rangle$, the equation of our new plane is:

$5x + 5y + 4z = D$

To find $D$, we use the point $P = (5, 5, 5)$ which lies on the plane. Substituting $P$ into the plane equation:

$5(5) + 5(5) + 4(5) = D$

Calculating the left side:

$25 + 25 + 20 = 70$

So, $D = 70$. Therefore, the equation of the plane passing through $P = (5, 5, 5)$ and parallel to the plane $5x + 5y + 4z = -6$ is:

$5x + 5y + 4z = 70$

Would you like further details or have any questions?

Here are 8 related questions that expand on this topic:

1. How do you determine the normal vector of a plane given its equation?
2. What is the general form of a plane equation in three dimensions?
3. How do you find the distance between a point and a plane?
4. What is the geometric significance of the normal vector of a plane?
5. How can you find the intersection line of two planes?
6. How do you determine if two planes are perpendicular?
7. How do you find the angle between a line and a plane?
8. How can you find the equation of a plane given three points on the plane?

Tip: When writing the equation of a plane, always ensure that the coefficients of $x$, $y$, and $z$ represent the components of the normal vector to the plane.