Write down an equation for the plane passing through the points [1,−1,−1][1,−1,−1], [−2,2,−1][−2,2,−1], and [−3,0,2][−3,0,2]. For example: x+y+z=1.



To find the equation of a plane passing through three given points, we need to determine a normal vector to the plane using the cross product of two vectors lying on the plane. Let's go through the step-by-step process.

### Given Points

- \( A = [1, -1, -1] \)
- \( B = [-2, 2, -1] \)
- \( C = [-3, 0, 2] \)

### Step 1: Calculate Vectors on the Plane

Let's find two vectors that lie on the plane by taking the difference between the given points:

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

#### Vector \(\mathbf{AC}\):
\[
\mathbf{AC} = C - A = [-3, 0, 2] - [1, -1, -1] = [-4, 1, 3]
\]

### Step 2: Find the Normal Vector \(\mathbf{n}\)

The normal vector \(\mathbf{n}\) to the plane can be found using the cross product of \(\mathbf{AB}\) and \(\mathbf{AC}\):

\[
\mathbf{n} = \mathbf{AB} \times \mathbf{AC} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-3 & 3 & 0 \\
-4 & 1 & 3
\end{vmatrix}
\]

#### Calculating the determinant:
\[
\mathbf{n} = \mathbf{i}(3 \cdot 3 - 0 \cdot 1) - \mathbf{j}(-3 \cdot 3 - 0 \cdot -4) + \mathbf{k}(-3 \cdot 1 - 3 \cdot -4)
\]

\[
= \mathbf{i}(9) - \mathbf{j}(-9) + \mathbf{k}(-3 + 12)
\]

\[
= \mathbf{i}(9) + \mathbf{j}(9) + \mathbf{k}(9)
\]

\[
= [9, 9, 9]
\]

**Simplifying**, \(\mathbf{n}\) becomes \([1, 1, 1]\).

### Step 3: Equation of the Plane

The general equation of a plane with normal vector \(\mathbf{n} = [a, b, c]\) passing through a point \(A(x_0, y_0, z_0)\) is:

\[
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0
\]

Using point \(A = [1, -1, -1]\) and \(\mathbf{n} = [1, 1, 1]\):

\[
1(x - 1) + 1(y + 1) + 1(z + 1) = 0
\]

Simplify this equation:

\[
(x - 1) + (y + 1) + (z + 1) = 0
\]

\[
x + y + z + 1 = 0
\]

\[
x + y + z = -1
\]

### Final Equation of the Plane

The equation of the plane passing through the points \([1, -1, -1]\), \([-2, 2, -1]\), and \([-3, 0, 2]\) is:

\[
\boxed{x + y + z = -1}
\]

### Verification

Let's verify that the plane equation satisfies all three given points.

#### Point \(A = [1, -1, -1]\):
\[
1 + (-1) + (-1) = -1 \quad \text{(True)}
\]

#### Point \(B = [-2, 2, -1]\):
\[
-2 + 2 + (-1) = -1 \quad \text{(True)}
\]

#### Point \(C = [-3, 0, 2]\):
\[
-3 + 0 + 2 = -1 \quad \text{(True)}
\]

Thus, the plane equation \(x + y + z = -1\) satisfies all three points, confirming it is correct.

---

If you have any further questions or need more clarification, feel free to ask! Here are some suggested questions you might consider exploring:

1. **How does the cross product help in determining the normal vector?**
2. **Can the equation of a plane be simplified further?**
3. **What if two of the given points are the same?**
4. **How to find the distance from a point to this plane?**
5. **What happens if the points are collinear?**

### Tip:

When determining the equation of a plane, ensure that your normal vector is non-zero; otherwise, the points might be collinear.

Learn how to find the equation of a plane passing through three points: [1,-1,-1], [-2,2,-1], and [-3,0,2]. Detailed step-by-step solution involving vectors, cross product, and plane equations. Suitable for Grades 10-12.

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