To find the equation of the plane that is perpendicular to the vector $\mathbf{n} = 3\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}$ and passing through the point $(-4, 4, 2)$, we use the general form of the plane equation:

$3x + 3y + 5z = D$

Here, $\mathbf{n}$ serves as the normal vector to the plane. To determine $D$, substitute the coordinates of the given point $(-4, 4, 2)$ into the plane equation.

Step 1: Substituting the point

$3(-4) + 3(4) + 5(2) = D$

$-12 + 12 + 10 = D$

$D = 10$

Step 2: Write the final equation

The equation of the plane is: $3x + 3y + 5z = 10$

Would you like me to explain this further or solve a related problem? Here are some follow-up questions for deeper exploration:

1. How do you verify if a point lies on this plane?
2. Can you find the distance from another point to this plane?
3. What is the intersection line of this plane with another plane?
4. How do you calculate the angle between this plane and another plane?
5. Can you rewrite this plane equation in parametric form?

Tip: The normal vector of a plane helps to define its orientation in space, and any point on the plane can be used to calculate its equation.