To solve the problem, we need to find the values of \( x \) and \( y \) given the vectors of triangle \( ABC \) and its centroid \( G \).
### Step 1: Understand the Centroid Formula
The centroid \( G \) of a triangle with vertices represented by vectors \( A \), \( B \), and \( C \) is given by the formula:
\[
G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)
\]
where \( (x_1, y_1, z_1) \), \( (x_2, y_2, z_2) \), and \( (x_3, y_3, z_3) \) are the coordinates of points \( A \), \( B \), and \( C \).
### Step 2: Identify Coordinates
The coordinates of points \( A \), \( B \), and \( C \) are given as:
- \( A = (3, x, 1) \)
- \( B = (y, -2, 2) \)
- \( C = (2x, 2y, -3) \)
The centroid \( G \) is given as \( (2, 1, 0) \).
### Step 3: Set Up the Equations
Using the centroid formula, we can set up three equations based on the x, y, and z coordinates.
1. For the x-coordinate:
\[
2 = \frac{3 + y + 2x}{3}
\]
Multiplying both sides by 3:
\[
6 = 3 + y + 2x \implies y + 2x = 3 \quad \text{(Equation 1)}
\]
2. For the y-coordinate:
\[
1 = \frac{x - 2 + 2y}{3}
\]
Multiplying both sides by 3:
\[
3 = x - 2 + 2y \implies x + 2y = 5 \quad \text{(Equation 2)}
\]
3. For the z-coordinate:
\[
0 = \frac{1 + 2 - 3}{3}
\]
This simplifies to:
\[
0 = 0 \quad \text{(This equation is always true and does not provide any new information)}
\]
### Step 4: Solve the System of Equations
Now we have a system of two equations:
1. \( y + 2x = 3 \)
2. \( x + 2y = 5 \)
We can solve these equations simultaneously.
From Equation 1, we can express \( y \) in terms of \( x \):
\[
y = 3 - 2x \quad \text{(Substituting into Equation 2)}
\]
Now substitute \( y \) into Equation 2:
\[
x + 2(3 - 2x) = 5
\]
Expanding this gives:
\[
x + 6 - 4x = 5
\]
Combining like terms:
\[
-3x + 6 = 5
\]
Subtracting 6 from both sides:
\[
-3x = -1 \implies x = \frac{1}{3}
\]
### Step 5: Find \( y \)
Now substitute \( x \) back into Equation 1 to find \( y \):
\[
y = 3 - 2\left(\frac{1}{3}\right) = 3 - \frac{2}{3} = \frac{9}{3} - \frac{2}{3} = \frac{7}{3}
\]
### Final Values
Thus, the values are:
\[
x = \frac{1}{3}, \quad y = \frac{7}{3}
\]
