Find the equation of variation for a point P (x, y, z) that moves so that it is always equidistant from the points A(1, 2, 3) and B(4, 6, 8)

Step 1 ：Given a point P(x, y, z) equidistant from points A(1, 2, 3) and B(4, 6, 8), the distance from P to A is equal to the distance from P to B. We can write this as \(AP = BP\).

Step 2 ：The distance between two points in 3D space is given by the formula \(D = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2 + (z_{2}-z_{1})^2}\).

Step 3 ：Substituting the given points into the distance formula, we get \(\sqrt{(x-1)^2 + (y-2)^2 + (z-3)^2} = \sqrt{(x-4)^2 + (y-6)^2 + (z-8)^2}\)

Step 4 ：Squaring both sides to eliminate the square root, we get \((x-1)^2 + (y-2)^2 + (z-3)^2 = (x-4)^2 + (y-6)^2 + (z-8)^2\)

Step 5 ：Expanding and simplifying the equation, we get \(3x + 4y + 6z = 21\).