The equation of the locus of point which are equidistance from \((4,5,2)\) and \((1,6,3)\) is (A) \(6 x-2 y-2 z+1=0\) (B) \(6 \mathrm{x}+2 \mathrm{y}-2 \mathrm{z}+1=0\) (C) \(6 \mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+1=0\) (D) \(6 \mathrm{x}-2 \mathrm{y}-2 \mathrm{z}-1=0\)

When dealing with geometry problems in three-dimensional space, we often need to calculate the distance between two points. The distance formula in 3D is an extension of the Pythagorean theorem into a third dimension, which makes it an essential tool for solving such problems.

The formula for calculating the distance, 'd', between two points, \(P_1=(x_1, y_1, z_1)\) and \(P_2=(x_2, y_2, z_2)\), is expressed as:
\[d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\]
This equation accounts for the horizontal, vertical, and 'depth' measurements between points, giving us the straight-line distance in 3D space. It's particularly useful in finding the locus of points equidistant from two fixed points, which lies on the perpendicular bisector and, in three dimensions, defines a plane.

Understanding the equation of a plane is crucial when we want to represent a flat, two-dimensional surface within 3D space. In analytic geometry, a plane can be defined by a linear equation in the form:
\[Ax + By + Cz + D = 0\]
where 'A', 'B', and 'C' are not all zero. This equation tells us that any point with coordinates \( (x, y, z) \) that satisfies this equation lies on the plane. The coefficients 'A', 'B', and 'C' represent the components of a vector that is perpendicular to the plane, known as the normal vector.

The locus of points that are equidistant from two given points forms a plane that bisects the segment joining the two points. By using the distance formula in 3D and algebraic manipulation, we can derive the specific equation for this plane, as was done in the initial exercise.

The ability to simplify algebraic equations is a cornerstone of solving many mathematical problems. Algabraic manipulation involves operations such as expanding brackets, collecting like terms, and isolating variables to simplify expressions or solve for unknowns.

In the context of our exercise, we started with a complex equation derived from equating the distances. To simplify this equation, we performed the following steps:

• Expanded the squared binomials.
• Grouped like terms for \(x\), \(y\), and \(z\).
• Combined and cancelled terms on both sides of the equation.
• Divided by common factors to reduce complexity.
• Multiplied by -1 to eliminate fractional coefficients.

Through these steps of simplification, we reduced our complex distance equation to the linear equation of a plane. This process demonstrates the power of algebraic simplification in making complex problems manageable and solvable.