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The distance formula is a fundamental concept in 3D geometry that allows us to calculate the straight-line distance between two points in space. In three dimensions, if you have two points, say \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \), the distance \( d \) between these points is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]. Understanding this formula helps to solve problems that involve finding how far apart two points are, useful in a variety of applications, from physics to computer graphics.

• Step-by-step: Subtract corresponding coordinates of the points.
• Square each difference.
• Add these squares.
• Finally, take the square root of the sum to find the distance.

This approach relies on the Pythagorean theorem extended into three dimensions, giving you a direct way to measure distances in space.

In 3D geometry, a sphere is the set of all points that are equidistant from a central point. The general equation for a sphere with center \( (h, k, l) \) and radius \( r \) is: \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\]. This form captures the idea that every point \((x, y, z)\) on the sphere is exactly \( r \) units away from the center \((h, k, l)\). For the problem at hand, we establish a sphere-form equation derived from given conditions about distances from certain fixed points. This involves aligning with the principle of balance in distances calculated through the previously understood formula.

• The squared distances from points to derive center and radius.
• Clear isolation of expressions related to \( x, y, \) and \( z \) variables.

Understanding the sphere's equation is essential for visualizing how far a point is in any direction from a central reference.

Completing the square is a technique employed for converting quadratic expressions into a form that easily reveals the properties essential for geometric interpretations, such as finding the center and radius of a sphere. The process involves adjusting a quadratic equation by adding and subtracting terms to transform it into a perfect square trinomial.For instance, consider a term like \( x^2 + bx \). To complete the square, follow these steps:

• Take half of the coefficient of \( x \), which is \( \frac{b}{2} \).
• Square this half to get \( \left(\frac{b}{2}\right)^2 \).
• Add and subtract this square inside the expression.
• Rewrite as a perfect square: \( (x + \frac{b}{2})^2 - \left(\frac{b}{2}\right)^2 \).

This rewriting enables the explicit formation of a sphere equation from the initial quadratic forms, preserving the mathematical integrity while simplifying analytical geometry tasks.

To determine the radius and center of a sphere from its equation, we look for a setup expressed in the standard form: \[(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\]. In this context, \( (h, k, l) \) denotes the center of the sphere, and \( r \) symbolizes the radius. To extract these values from an expanded equation:

• Complete the square for all three variables, \( x, y, \) and \( z \), organizing them into the square form.
• Identify the constants within each squared term to pinpoint the center \((h, k, l)\).
• Calculate \( r \) by rearranging the equation to its standard form, equivalent to the constant on the sphere-side post equal sign.

For example, given in this problem, the process involved reducing a complex equation into one where the factors align with the form \( (x - \frac{25}{3})^2 + (y - \frac{7}{3})^2 + (z + \frac{11}{3})^2 = \left(\frac{\sqrt{499}}{3}\right)^2 \). This shows the sphere's center and simplifies finding the radius as \( \frac{\sqrt{499}}{3} \). Using these techniques consistently will assist in effectively interpreting and solving complex geometry problems.