The equation of a plane parallel to another plane has the same normal vector as the given plane. The general equation of a plane is:

$Ax + By + Cz = D$

Where $A$, $B$, and $C$ are the components of the normal vector.

Step 1: Identify the normal vector of the given plane

The equation of the given plane is:

$5x - 5y - z = -6$

The normal vector $\mathbf{n}$ of this plane is the coefficients of $x$, $y$, and $z$, which gives:

$\mathbf{n} = (5, -5, -1)$

Step 2: Write the general equation of the plane parallel to the given one

Since the plane we're looking for is parallel to the given plane, it will have the same normal vector, so its equation will have the form:

$5x - 5y - z = D$

Step 3: Use the point $P(3, 3, 2)$ to find $D$

Substitute the coordinates of point $P(3, 3, 2)$ into the equation to find $D$:

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

$15 - 15 - 2 = D$

$D = -2$

Step 4: Final equation of the plane

Therefore, the equation of the plane through the point $P(3, 3, 2)$ and parallel to the plane $5x - 5y - z = -6$ is:

$5x - 5y - z = -2$

Would you like more details on this or have any further questions?

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Here are 5 related questions you might consider:

1. How do you determine the normal vector from a plane equation?
2. What is the geometric interpretation of parallel planes?
3. Can you find the distance between two parallel planes?
4. How do you find the equation of a plane passing through three points?
5. What is the equation of a plane perpendicular to a given vector?

Tip: The constant $D$ in the equation of a plane can be quickly determined by substituting any known point on the plane into the equation.