Question:

Step 1: Determine the midpoint of the line segment. The given points are $A(-3, -3, 4)$ and $B(3, 7, 6)$. The midpoint $M$ of the line segment $AB$ is calculated as: $$M = \left(\frac{-3+3}{2}, \frac{-3+7}{2}, \frac{4+6}{2}\right) = (0, 2, 5)$$ Step 2: Determine the normal vector to the plane. The plane bisects the line segment at right angles, so its normal vector is parallel to the direction vector of the segment. The direction vector $\vec{AB}$ is: $$\vec{AB} = B - A = (3 - (-3), 7 - (-3), 6 - 4) = (6, 10, 2)$$ A simplified normal vector $\vec{n}$ can be obtained by dividing by 2: $$\vec{n} = (3, 5, 1)$$ Step 3: Write the equation of the perpendicular bisecting plane. The plane passes through the midpoint $M(0, 2, 5)$ and has a normal vector $\vec{n} = (3, 5, 1)$. The equation of the plane is given by $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$: $$3(x - 0) + 5(y - 2) + 1(z - 5) = 0$$ $$3x + 5y - 10 + z - 5 = 0$$ $$3x + 5y + z = 15$$ Step 4: Identify the point that lies on the plane. A point $(x, y, z)$ lies on the plane if it satisfies the equation $3x + 5y + z = 15$. For the point $(-2, 3, 5)$: $$3(-2) + 5(3) + 5 = -6 + 15 + 5 = 14$$ Since $14 \neq 15$, the point $(-2, 3, 5)$ does not lie on the plane. For the point $(4, 1, -2)$: $$3(4) + 5(1) + (-2) = 12 + 5 - 2 = 15$$ Since $15 = 15$, the point $(4, 1, -2)$ lies on the plane.