Key Concepts and Formulas

• Midpoint Formula: The coordinates of the midpoint $M$ of a line segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)$
• Equation of a Plane: The equation of a plane passing through a point $(x_0, y_0, z_0)$ and having a normal vector with direction ratios $(A, B, C)$ is given by: $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$
• Perpendicular Bisector Plane: A plane that bisects a line segment at right angles (a perpendicular bisector plane) has two key properties:
  1. It passes through the midpoint of the line segment.
  2. Its normal vector is parallel to the line segment itself. This means the direction ratios of the line segment are the direction ratios of the plane's normal vector.

Step-by-Step Solution

Let the two given points be $P_1 = (1, 2, 3)$ and $P_2 = (-3, 4, 5)$.

Step 1: Determine the Midpoint of the Line Segment

• Why this step? The problem states that the plane "bisects" the line segment. For a plane to bisect a segment, it must pass through the exact middle point of that segment. Therefore, we first find the midpoint, which will be a point lying on the required plane.
• Using the midpoint formula for $P_1(1, 2, 3)$ and $P_2(-3, 4, 5)$: $M = \left( \frac{1 + (-3)}{2}, \frac{2 + 4}{2}, \frac{3 + 5}{2} \right)$ $M = \left( \frac{-2}{2}, \frac{6}{2}, \frac{8}{2} \right)$ $M = (-1, 3, 4)$
• Thus, the required plane passes through the point $(-1, 3, 4)$.

Step 2: Determine the Direction Ratios of the Normal Vector to the Plane

• Why this step? The problem states that the plane bisects the line segment "at right angles". This means the line segment $P_1P_2$ is perpendicular to the plane. If a line is perpendicular to a plane, then its direction ratios are identical to the direction ratios of the normal vector to that plane.
• The direction ratios of the line segment $P_1P_2$ joining $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.
• Using $P_1(1, 2, 3)$ and $P_2(-3, 4, 5)$, the direction ratios of the normal vector $(A, B, C)$ to the plane are: $A = x_2 - x_1 = -3 - 1 = -4$ $B = y_2 - y_1 = 4 - 2 = 2$ $C = z_2 - z_1 = 5 - 3 = 2$
• Therefore, the direction ratios of the normal vector to the plane are $(-4, 2, 2)$. We can simplify these direction ratios by dividing by a common factor of 2, resulting in $(-2, 1, 1)$, which will lead to an equivalent plane equation. We will use $(-4, 2, 2)$ for now and simplify the final equation.

Step 3: Formulate the Equation of the Plane

• Why this step? We now have all the necessary information to write the equation of the plane: a point on the plane $M(-1, 3, 4)$ (from Step 1) and the direction ratios of its normal vector $(-4, 2, 2)$ (from Step 2).
• Using the general equation of a plane $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$, with $(x_0, y_0, z_0) = (-1, 3, 4)$ and $(A, B, C) = (-4, 2, 2)$: $-4(x - (-1)) + 2(y - 3) + 2(z - 4) = 0$ $-4(x + 1) + 2(y - 3) + 2(z - 4) = 0$
• Expand and simplify the equation: $-4x - 4 + 2y - 6 + 2z - 8 = 0$ $-4x + 2y + 2z - 18 = 0$ $-4x + 2y + 2z = 18$
• To simplify the coefficients, we can divide the entire equation by 2: $-2x + y + z = 9$ This is the equation of the required plane.

Step 4: Check Which Option Point Lies on the Plane

• Why this step? The question asks for a point from the given options that also lies on this plane. To check if a point lies on a plane, we substitute its coordinates into the plane's equation. If the equation holds true (left-hand side equals right-hand side), then the point lies on the plane.

• Let's test each option using the plane equation $-2x + y + z = 9$:

  • (A) (-3, 2, 1): Substitute $x = -3, y = 2, z = 1$: $-2(-3) + (2) + (1) = 6 + 2 + 1 = 9$ Since $9 = 9$, the point $(-3, 2, 1)$ lies on the plane.

  • (B) (3, 2, 1): Substitute $x = 3, y = 2, z = 1$: $-2(3) + (2) + (1) = -6 + 2 + 1 = -3$ Since $-3 \neq 9$, the point $(3, 2, 1)$ does not lie on the plane.

  • (C) (-1, 2, 3): Substitute $x = -1, y = 2, z = 3$: $-2(-1) + (2) + (3) = 2 + 2 + 3 = 7$ Since $7 \neq 9$, the point $(-1, 2, 3)$ does not lie on the plane.

  • (D) (1, 2, -3): Substitute $x = 1, y = 2, z = -3$: $-2(1) + (2) + (-3) = -2 + 2 - 3 = -3$ Since $-3 \neq 9$, the point $(1, 2, -3)$ does not lie on the plane.

• Only option (A) satisfies the equation of the plane.

Common Mistakes & Tips

• Forgetting the "Right Angles" Condition: A common error is just finding the midpoint and assuming the plane passes through it, without using the perpendicularity condition to find the normal vector. Remember, "at right angles" implies the line segment is perpendicular to the plane.
• Direction Ratios vs. Normal Vector: Ensure you correctly identify the direction ratios of the line segment as the normal vector's direction ratios for a perpendicular plane. If the plane were parallel to the line, a different approach for the normal vector would be needed.
• Algebraic Errors: Be careful with signs when calculating midpoints, direction ratios, and expanding the plane equation. Double-check your arithmetic, especially when substituting values.

Summary

To find the plane that bisects a line segment at right angles, we first determine the midpoint of the segment, as the plane must pass through it. Then, we use the fact that the plane is perpendicular to the segment, meaning the direction ratios of the segment serve as the direction ratios of the plane's normal vector. With a point on the plane and its normal vector, we can formulate the equation of the plane. Finally, we test the given options to find the point that satisfies the plane's equation. Following these steps, we found the plane equation to be $-2x + y + z = 9$, and option (A) $(-3, 2, 1)$ satisfies this equation.

The final answer is $\boxed{(-3, 2, 1)}$, which corresponds to option (A).