The equation of the plane passing through the point
`(-2,-2,2)` and containing the line joining the points `(1,1,1)` and `(1,-1,2)` is

Options

A. `x+2y-3z+4=0`
B. `3x-4y+1=0`
C. `5x+2y-3z+1=0`
D. `x-3y-6z+8=0`

Answer

To find the equation of the plane that passes through the point \((-2, -2, 2)\) and contains the line joining the points \((1, 1, 1)\) and \((1, -1, 2)\), we can follow these steps:

### Step 1: Identify the Points
Let:
- Point \( A = (-2, -2, 2) \)
- Point \( B = (1, 1, 1) \)
- Point \( C = (1, -1, 2) \)

### Step 2: Find Vectors \( \overrightarrow{AC} \) and \( \overrightarrow{BC} \)
1. **Calculate vector \( \overrightarrow{AC} \)**:
\[ \overrightarrow{AC} = C - A = (1, -1, 2) - (-2, -2, 2) = (1 + 2, -1 + 2, 2 - 2) = (3, 1, 0) \]

2. **Calculate vector \( \overrightarrow{BC} \)**:
\[ \overrightarrow{BC} = C - B = (1, -1, 2) - (1, 1, 1) = (1 - 1, -1 - 1, 2 - 1) = (0, -2, 1) \]

### Step 3: Find the Normal Vector of the Plane
The normal vector \( \mathbf{n} \) to the plane can be found using the cross product of vectors \( \overrightarrow{AC} \) and \( \overrightarrow{BC} \):
\[ \mathbf{n} = \overrightarrow{AC} \times \overrightarrow{BC} \]

Using the determinant method:
\[ \mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 0 \\ 0 & -2 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ \mathbf{n} = \mathbf{i}(1 \cdot 1 - 0 \cdot -2) - \mathbf{j}(3 \cdot 1 - 0 \cdot 0) + \mathbf{k}(3 \cdot -2 - 1 \cdot 0) \]
\[ = \mathbf{i}(1) - \mathbf{j}(3) + \mathbf{k}(-6) \]
\[ = (1, -3, -6) \]

### Step 4: Write the Equation of the Plane
The general equation of a plane is given by:
\[ n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0 \]
where \((x_0, y_0, z_0)\) is a point on the plane and \((n_x, n_y, n_z)\) is the normal vector.

Substituting the values:
\[ 1(x + 2) - 3(y + 2) - 6(z - 2) = 0 \]

Expanding this:
\[ x + 2 - 3y - 6 - 6z + 12 = 0 \]
\[ x - 3y - 6z + 8 = 0 \]

### Final Equation of the Plane
Thus, the equation of the plane is:
\[ x - 3y - 6z + 8 = 0 \]