Find the equation of the plane through the points A(3, 2, 0), B(1, 3, -1) and C(0, -2, 3).

Answer

To find the equation of the plane through the points A(3, 2, 0), B(1, 3, -1), and C(0, -2, 3), we can follow these steps:

### Step 1: Find the vectors AB and AC
First, we need to find the vectors AB and AC using the coordinates of points A, B, and C.

- **Vector AB**:
\[ \text{AB} = \text{B} - \text{A} = (1 - 3, 3 - 2, -1 - 0) = (-2, 1, -1) \]

- **Vector AC**:
\[ \text{AC} = \text{C} - \text{A} = (0 - 3, -2 - 2, 3 - 0) = (-3, -4, 3) \]

### Step 2: Find the normal vector to the plane
The normal vector \( \mathbf{n} \) to the plane can be found by taking the cross product of vectors AB and AC.

\[ \mathbf{n} = \mathbf{AB} \times \mathbf{AC} \]

Using the determinant form for the cross product:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & -1 \\ -3 & -4 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{n} = \mathbf{i} \left(1 \cdot 3 - (-1)(-4)\right) - \mathbf{j} \left(-2 \cdot 3 - (-1)(-3)\right) + \mathbf{k} \left(-2 \cdot (-4) - 1 \cdot (-3)\right) \]

Calculating each component:
- For \( \mathbf{i} \): \( 3 - 4 = -1 \)
- For \( \mathbf{j} \): \( -6 - 3 = -9 \) (note the negative sign in front)
- For \( \mathbf{k} \): \( 8 + 3 = 11 \)

Thus, the normal vector \( \mathbf{n} \) is:
\[ \mathbf{n} = (-1, 9, 11) \]

### Step 3: Write the equation of the plane
The equation of the plane can be expressed in the form:
\[ \mathbf{n} \cdot (\mathbf{r} - \mathbf{a}) = 0 \]
where \( \mathbf{a} \) is a point on the plane (we can use point A) and \( \mathbf{r} = (x, y, z) \).

Substituting the normal vector and point A into the equation:
\[ (-1, 9, 11) \cdot ((x, y, z) - (3, 2, 0)) = 0 \]

Expanding this:
\[ -1(x - 3) + 9(y - 2) + 11(z - 0) = 0 \]

Simplifying:
\[ -x + 3 + 9y - 18 + 11z = 0 \]
\[ -x + 9y + 11z - 15 = 0 \]

Rearranging gives the equation of the plane:
\[ x - 9y - 11z + 15 = 0 \]

### Final Answer
The equation of the plane through the points A, B, and C is:
\[ x - 9y - 11z + 15 = 0 \]