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To solve the problem of finding the distance from the point \((-3, 1, 0)\) to the plane \(-3x + 3y + 3z = 9\), we use the formula for the distance between a point \((x_1, y_1, z_1)\) and a plane \(Ax + By + Cz + D = 0\):

\[
\text{Distance} = \frac{|A x_1 + B y_1 + C z_1 + D|}{\sqrt{A^2 + B^2 + C^2}}
\]

### Step 1: Rewrite the plane equation
The given equation is:
\[
-3x + 3y + 3z = 9
\]
Rewrite it in standard form \(Ax + By + Cz + D = 0\):
\[
-3x + 3y + 3z - 9 = 0
\]
Here, \(A = -3\), \(B = 3\), \(C = 3\), and \(D = -9\).

### Step 2: Substitute the point into the formula
The coordinates of the point are \((-3, 1, 0)\), so \(x_1 = -3\), \(y_1 = 1\), and \(z_1 = 0\). Plugging these into the distance formula:

\[
\text{Distance} = \frac{|(-3)(-3) + (3)(1) + (3)(0) - 9|}{\sqrt{(-3)^2 + 3^2 + 3^2}}
\]

### Step 3: Simplify the numerator
\[
\text{Numerator} = |(-3)(-3) + (3)(1) + (3)(0) - 9|
= |9 + 3 + 0 - 9| = |3|
\]

### Step 4: Simplify the denominator
\[
\text{Denominator} = \sqrt{(-3)^2 + 3^2 + 3^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}
\]

### Step 5: Compute the distance
\[
\text{Distance} = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
\]

### Final Answer:
The distance from the point \((-3, 1, 0)\) to the plane \(-3x + 3y + 3z = 9\) is:
\[
\boxed{\frac{\sqrt{3}}{3}}
\]

Would you like further clarification or steps explained in detail?

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Here are 5 related questions to explore:
1. What is the geometric meaning of the distance between a point and a plane?
2. How do you find the equation of a plane given three points?
3. Can the distance formula be applied in higher dimensions (e.g., 4D space)?
4. How is the distance formula derived from vector projections?
5. What changes if the plane equation is written differently (e.g., with coefficients scaled)?

### Tip:
To simplify calculations, always ensure the plane equation is written in standard form \(Ax + By + Cz + D = 0\).

Find the distance from the point (-3, 1, 0) to the plane -3x + 3y + 3z = 9.

Learn how to calculate the distance from the point (-3, 1, 0) to the plane -3x + 3y + 3z = 9 using the point-to-plane distance formula. A detailed step-by-step solution explains the process, suitable for high school students studying geometry and vectors.

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