Find the equation of the set of points which are equidistant from the points `(1,2,3)` and `(3,2,-1)`

Answer

To find the equation of the set of points that are equidistant from the points \( A(1, 2, 3) \) and \( B(3, 2, -1) \), we can follow these steps:

### Step 1: Define the point P
Let \( P(x, y, z) \) be a point that is equidistant from points \( A \) and \( B \).

### Step 2: Set up the distance equations
The distance from point \( P \) to point \( A \) is given by:
\[ PA = \sqrt{(x - 1)^2 + (y - 2)^2 + (z - 3)^2} \]
The distance from point \( P \) to point \( B \) is given by:
\[ PB = \sqrt{(x - 3)^2 + (y - 2)^2 + (z + 1)^2} \]

### Step 3: Set the distances equal
Since \( P \) is equidistant from \( A \) and \( B \), we have:
\[ PA = PB \]
Squaring both sides to eliminate the square roots gives:
\[ (x - 1)^2 + (y - 2)^2 + (z - 3)^2 = (x - 3)^2 + (y - 2)^2 + (z + 1)^2 \]

### Step 4: Expand both sides
Expanding the left side:
\[ (x - 1)^2 = x^2 - 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
\[ (z - 3)^2 = z^2 - 6z + 9 \]
So,
\[ x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 \]
This simplifies to:
\[ x^2 + y^2 + z^2 - 2x - 4y - 6z + 14 \]

Now expanding the right side:
\[ (x - 3)^2 = x^2 - 6x + 9 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
\[ (z + 1)^2 = z^2 + 2z + 1 \]
So,
\[ x^2 - 6x + 9 + y^2 - 4y + 4 + z^2 + 2z + 1 \]
This simplifies to:
\[ x^2 + y^2 + z^2 - 6x - 4y + 2z + 14 \]

### Step 5: Set the expanded equations equal
Now we have:
\[ x^2 + y^2 + z^2 - 2x - 4y - 6z + 14 = x^2 + y^2 + z^2 - 6x - 4y + 2z + 14 \]

### Step 6: Cancel like terms
Cancel \( x^2 \), \( y^2 \), \( z^2 \), and \( 14 \) from both sides:
\[ -2x - 6z = -6x + 2z \]

### Step 7: Rearrange the equation
Rearranging gives:
\[ -2x + 6x - 6z - 2z = 0 \]
This simplifies to:
\[ 4x - 8z = 0 \]

### Step 8: Factor out common terms
Factoring out \( 4 \):
\[ 4(x - 2z) = 0 \]

### Final Equation
Thus, the equation of the set of points that are equidistant from points \( A \) and \( B \) is:
\[ x - 2z = 0 \]
or equivalently,
\[ x = 2z \]