A plane passes through the point `(-2, -2, 2)` and contains the line joining the points `(1, -1, 2) and (1, 1, 1).` Then the image of `(-7, 2, 3)` in the plane is

Options

A. `(1, -1, 5)`
B. `(-5, -4,-2)`
C. `(-6, -1, -3)`
D. `((13)/(23),(7)/(23),(6)/(23))`

Answer

To solve the problem step by step, we will find the equation of the plane that contains the given points and then calculate the image of the point \((-7, 2, 3)\) in that plane.

### Step 1: Find the direction vectors of the line segment BC and AC.

1. **Identify the points**:
- Point A: \((-2, -2, 2)\)
- Point B: \((1, -1, 2)\)
- Point C: \((1, 1, 1)\)

2. **Calculate the vectors**:
- Vector **AC** = C - A = \((1 - (-2), 1 - (-2), 1 - 2) = (3, 3, -1)\)
- Vector **BC** = C - B = \((1 - 1, 1 - (-1), 1 - 2) = (0, 2, -1)\)

### Step 2: Find the normal vector of the plane.

3. **Compute the cross product of AC and BC**:
\[ \text{Normal vector } \mathbf{n} = \mathbf{AC} \times \mathbf{BC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 3 & -1 \\ 0 & 2 & -1 \end{vmatrix} \]
- Calculate the determinant:
\[ \mathbf{n} = \mathbf{i}(3 \cdot (-1) - (-1) \cdot 2) - \mathbf{j}(3 \cdot (-1) - (-1) \cdot 0) + \mathbf{k}(3 \cdot 2 - 3 \cdot 0) \]
\[ = \mathbf{i}(-3 + 2) - \mathbf{j}(-3) + \mathbf{k}(6) \]
\[ = (-1, 3, 6) \]

### Step 3: Write the equation of the plane.

4. **Use the point-normal form of the plane equation**:
The equation of the plane can be expressed as:
\[ n_x(x - x_0) + n_y(y - y_0) + n_z(z - z_0) = 0 \]
Where \((x_0, y_0, z_0)\) is point A \((-2, -2, 2)\) and \((n_x, n_y, n_z) = (-1, 3, 6)\):
\[ -1(x + 2) + 3(y + 2) + 6(z - 2) = 0 \]
Simplifying this:
\[ -x - 2 + 3y + 6 + 6z - 12 = 0 \]
\[ -x + 3y + 6z - 8 = 0 \]
Rearranging gives:
\[ x - 3y - 6z + 8 = 0 \]

### Step 4: Find the image of the point \((-7, 2, 3)\) in the plane.

5. **Let the point P be \((-7, 2, 3)\)** and find the equation of the line perpendicular to the plane passing through P:
- The direction ratios of the line are given by the normal vector \((-1, 3, 6)\).
- The parametric equations of the line are:
\[ x = -7 - t, \quad y = 2 + 3t, \quad z = 3 + 6t \]

6. **Find the intersection of this line with the plane**:
Substitute \(x, y, z\) into the plane equation:
\[ (-7 - t) - 3(2 + 3t) - 6(3 + 6t) + 8 = 0 \]
Simplifying:
\[ -7 - t - 6 - 9t - 18 - 36t + 8 = 0 \]
\[ -t - 9t - 36t - 23 = 0 \]
\[ -46t - 23 = 0 \implies t = -\frac{23}{46} = -\frac{1}{2} \]

7. **Find the coordinates of the intersection point Q**:
Substitute \(t = -\frac{1}{2}\) back into the parametric equations:
\[ x = -7 + \frac{1}{2} = -\frac{13}{2}, \quad y = 2 - \frac{3}{2} = \frac{1}{2}, \quad z = 3 - 3 = 0 \]
Thus, \(Q = \left(-\frac{13}{2}, \frac{1}{2}, 0\right)\).

8. **Find the image point R**:
The image point R is such that Q is the midpoint of P and R. Therefore:
\[ R = (2Q - P) = \left(2 \left(-\frac{13}{2}\right) + 7, 2 \left(\frac{1}{2}\right) - 2, 2(0) - 3\right) \]
\[ R = (-13 + 7, 1 - 2, 0 - 3) = (-6, -1, -3) \]

### Final Answer:
The image of the point \((-7, 2, 3)\) in the plane is \((-6, -1, -3)\).

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