1. Key Concepts and Formulas

• Midpoint of a Line Segment: If a point $P(x_1, y_1, z_1)$ has its mirror image $P'(x_2, y_2, z_2)$ in a plane, then the midpoint $M$ of the line segment $PP'$ always lies on the plane. The coordinates of the midpoint are given by: $M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} \right)$
• Normal Vector of the Plane: The line segment $PP'$ is perpendicular (normal) to the plane of reflection. Therefore, the direction ratios (DRs) of the line segment $PP'$ serve as the direction ratios of the normal vector to the plane. If $P=(x_1, y_1, z_1)$ and $P'=(x_2, y_2, z_2)$, the DRs of $PP'$ are $(x_2-x_1, y_2-y_1, z_2-z_1)$.
• Equation of a Plane: The equation of a plane passing through a point $(x_0, y_0, z_0)$ with a normal vector $\vec{n} = (a, b, c)$ is given by: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$

2. Step-by-Step Solution

Step 1: Identify the given points.

• What we are doing: We are explicitly noting down the coordinates of the original point and its mirror image.
• Why this step: This sets up our variables for subsequent calculations. Let the given point be $P = (1, 2, 3)$. Let its mirror image in the plane be $P' = \left( - \frac{7}{3}, - \frac{4}{3}, - \frac{1}{3} \right)$.

Step 2: Find the midpoint of the line segment $PP'$.

• What we are doing: Calculating the coordinates of the midpoint $M$ of $PP'$.
• Why this step: As per the key concepts, this midpoint $M$ must lie on the plane. This provides us with a specific point $(x_0, y_0, z_0)$ that the plane passes through.
• Applying the midpoint formula: $M = \left( \frac{1 + \left( - \frac{7}{3} \right)}{2}, \frac{2 + \left( - \frac{4}{3} \right)}{2}, \frac{3 + \left( - \frac{1}{3} \right)}{2} \right)$ Let's calculate each coordinate:
  • x-coordinate: $\frac{1 - \frac{7}{3}}{2} = \frac{\frac{3-7}{3}}{2} = \frac{-\frac{4}{3}}{2} = - \frac{4}{6} = - \frac{2}{3}$
  • y-coordinate: $\frac{2 - \frac{4}{3}}{2} = \frac{\frac{6-4}{3}}{2} = \frac{\frac{2}{3}}{2} = \frac{2}{6} = \frac{1}{3}$
  • z-coordinate: $\frac{3 - \frac{1}{3}}{2} = \frac{\frac{9-1}{3}}{2} = \frac{\frac{8}{3}}{2} = \frac{8}{6} = \frac{4}{3}$ So, the midpoint is $M = \left( - \frac{2}{3}, \frac{1}{3}, \frac{4}{3} \right)$. This point lies on the plane.

Step 3: Determine the Direction Ratios (DRs) of the normal vector to the plane.

• What we are doing: Finding the direction ratios of the line segment $PP'$.
• Why this step: The line segment connecting a point to its mirror image is always perpendicular to the plane of reflection. Therefore, the DRs of $PP'$ are the DRs of the plane's normal vector $(a, b, c)$.
• The direction ratios of the line segment $PP'$ are found by subtracting the coordinates of $P$ from $P'$: DRs of $PP' = \left( - \frac{7}{3} - 1, - \frac{4}{3} - 2, - \frac{1}{3} - 3 \right)$
  • x-component: $- \frac{7}{3} - \frac{3}{3} = - \frac{10}{3}$
  • y-component: $- \frac{4}{3} - \frac{6}{3} = - \frac{10}{3}$
  • z-component: $- \frac{1}{3} - \frac{9}{3} = - \frac{10}{3}$ The DRs of the normal vector are proportional to $\left( - \frac{10}{3}, - \frac{10}{3}, - \frac{10}{3} \right)$. We can simplify these ratios by multiplying by $-\frac{3}{10}$ (or dividing by $-\frac{10}{3}$) to get the simplest integer DRs: $(1, 1, 1)$. So, we can take the normal vector $\vec{n} = (a, b, c) = (1, 1, 1)$.

Step 4: Formulate the equation of the plane.

• What we are doing: Using the midpoint $M$ (a point on the plane) and the simplified normal vector DRs to write the plane's equation.
• Why this step: We now have all the necessary information to uniquely define the plane: a point it passes through and its orientation (normal vector).
• Using the general equation of a plane $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$, with $(x_0, y_0, z_0) = M = \left( - \frac{2}{3}, \frac{1}{3}, \frac{4}{3} \right)$ and $(a, b, c) = (1, 1, 1)$: $1 \left( x - \left( - \frac{2}{3} \right) \right) + 1 \left( y - \frac{1}{3} \right) + 1 \left( z - \frac{4}{3} \right) = 0$ $x + \frac{2}{3} + y - \frac{1}{3} + z - \frac{4}{3} = 0$ Combine the constant terms: $x + y + z + \left( \frac{2 - 1 - 4}{3} \right) = 0$ $x + y + z + \left( \frac{-3}{3} \right) = 0$ $x + y + z - 1 = 0$ Thus, the equation of the plane is $x + y + z = 1$.

Step 5: Check which of the given points lies on this plane.

• What we are doing: Substituting the coordinates of each option into the derived plane equation.
• Why this step: A point lies on the plane if its coordinates satisfy the plane's equation. We need to find the option that makes the equation true.
• The equation of the plane is $x + y + z = 1$.
  • (A) (1, –1, 1): Substitute $x=1, y=-1, z=1 \implies 1 + (-1) + 1 = 1$. This is true ($1 = 1$).
  • (B) (–1, –1, –1): Substitute $x=-1, y=-1, z=-1 \implies -1 + (-1) + (-1) = -3$. This is not equal to $1$.
  • (C) (–1, –1, 1): Substitute $x=-1, y=-1, z=1 \implies -1 + (-1) + 1 = -1$. This is not equal to $1$.
  • (D) (1, 1, 1): Substitute $x=1, y=1, z=1 \implies 1 + 1 + 1 = 3$. This is not equal to $1$.

Therefore, point (A) lies on the plane.

3. Common Mistakes & Tips

• Fraction Arithmetic: Be meticulous with calculations involving fractions, especially when finding the midpoint coordinates and direction ratios. A small error can propagate.
• Simplifying DRs: Always simplify the direction ratios of the normal vector to the smallest possible integers. This makes the coefficients in the plane equation simpler and reduces the chance of calculation errors.
• Conceptual Clarity: Remember that the midpoint of the segment $PP'$ lies on the plane, and the vector $PP'$ is normal to the plane. Do not confuse these roles.

4. Summary

To determine the equation of a plane given a point and its mirror image, we leverage two fundamental geometric properties. First, the midpoint of the line segment connecting the point and its image lies on the plane. Second, this line segment is perpendicular to the plane, providing the direction ratios of the plane's normal vector. By calculating these two components, we can construct the plane's equation. Finally, we test the given options by substituting their coordinates into the plane equation to find the point that satisfies it. Following these steps, we found the plane equation to be $x+y+z=1$, and the point $(1, -1, 1)$ satisfies this equation.

The final answer is \boxed{(1, -1, 1)} which corresponds to option (A).