To find which of the given points lies on the plane defined by the image of the point `(1, 2, 3)` with respect to the plane being `(-7/3, -4/3, -1/3)`, we can follow these steps:
### Step 1: Find the Midpoint
The midpoint of the original point `(1, 2, 3)` and its image `(-7/3, -4/3, -1/3)` will lie on the plane. The midpoint \( M \) can be calculated as follows:
\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)
\]
Substituting the coordinates:
\[
M = \left( \frac{1 + \left(-\frac{7}{3}\right)}{2}, \frac{2 + \left(-\frac{4}{3}\right)}{2}, \frac{3 + \left(-\frac{1}{3}\right)}{2} \right)
\]
Calculating each component:
1. \( x \)-coordinate:
\[
\frac{1 - \frac{7}{3}}{2} = \frac{\frac{3}{3} - \frac{7}{3}}{2} = \frac{-\frac{4}{3}}{2} = -\frac{2}{3}
\]
2. \( y \)-coordinate:
\[
\frac{2 - \frac{4}{3}}{2} = \frac{\frac{6}{3} - \frac{4}{3}}{2} = \frac{\frac{2}{3}}{2} = \frac{1}{3}
\]
3. \( z \)-coordinate:
\[
\frac{3 - \frac{1}{3}}{2} = \frac{\frac{9}{3} - \frac{1}{3}}{2} = \frac{\frac{8}{3}}{2} = \frac{4}{3}
\]
Thus, the midpoint \( M \) is:
\[
M = \left(-\frac{2}{3}, \frac{1}{3}, \frac{4}{3}\right)
\]
### Step 2: Find the Direction Ratios of the Normal to the Plane
The direction ratios \( A, B, C \) of the normal to the plane can be found from the difference of the coordinates of the original point and its image:
1. \( A = -\frac{7}{3} - 1 = -\frac{10}{3} \)
2. \( B = -\frac{4}{3} - 2 = -\frac{10}{3} \)
3. \( C = -\frac{1}{3} - 3 = -\frac{10}{3} \)
### Step 3: Equation of the Plane
Using the point-normal form of the plane equation:
\[
A(x - x_1) + B(y - y_1) + C(z - z_1) = 0
\]
Substituting \( (x_1, y_1, z_1) = \left(-\frac{2}{3}, \frac{1}{3}, \frac{4}{3}\right) \):
\[
-\frac{10}{3}\left(x + \frac{2}{3}\right) - \frac{10}{3}\left(y - \frac{1}{3}\right) - \frac{10}{3}\left(z - \frac{4}{3}\right) = 0
\]
Simplifying gives:
\[
x + y + z = 1
\]
### Step 4: Check Which Points Lie on the Plane
Now we check each option to see if they satisfy the equation \( x + y + z = 1 \):
1. **Option A: (-1, 1, -1)**
\[
-1 + 1 - 1 = -1 \quad (\text{not } 1)
\]
2. **Option B: (-1, -1, 1)**
\[
-1 - 1 + 1 = -1 \quad (\text{not } 1)
\]
3. **Option C: (-1, -1, 1)**
\[
-1 - 1 + 1 = -1 \quad (\text{not } 1)
\]
4. **Option D: (1, 1, -1)**
\[
1 + 1 - 1 = 1 \quad (\text{is } 1)
\]
### Conclusion
The point that lies on the plane is **Option D: (1, 1, -1)**.
