The locus of the point which is equidistant from the point `A(0, 2, 3) and B(2, -2, 1)` is

Options

A. x - 2y - z + 1 = 0
B. x + 2y - z - 1 = 0
C. x + 2y + z + 1 = 0
D. None of these

Answer

To find the locus of the point that is equidistant from points A(0, 2, 3) and B(2, -2, 1), we will follow these steps:

### Step 1: Define the point P
Let the coordinates of point P be \( P(x, y, z) \).

### Step 2: Write the distance formulas
The distance from point P to point A is given by:
\[ PA = \sqrt{(x - 0)^2 + (y - 2)^2 + (z - 3)^2} \]
The distance from point P to point B is given by:
\[ PB = \sqrt{(x - 2)^2 + (y + 2)^2 + (z - 1)^2} \]

### Step 3: Set the distances equal
Since point P is equidistant from points A and B, we have:
\[ PA = PB \]
Squaring both sides to eliminate the square roots gives:
\[ (x - 0)^2 + (y - 2)^2 + (z - 3)^2 = (x - 2)^2 + (y + 2)^2 + (z - 1)^2 \]

### Step 4: Expand both sides
Expanding the left side:
\[ x^2 + (y - 2)^2 + (z - 3)^2 = x^2 + (y^2 - 4y + 4) + (z^2 - 6z + 9) \]
This simplifies to:
\[ x^2 + y^2 - 4y + 4 + z^2 - 6z + 9 \]
So, the left side becomes:
\[ x^2 + y^2 + z^2 - 4y - 6z + 13 \]

Expanding the right side:
\[ (x - 2)^2 + (y + 2)^2 + (z - 1)^2 = (x^2 - 4x + 4) + (y^2 + 4y + 4) + (z^2 - 2z + 1) \]
This simplifies to:
\[ x^2 - 4x + 4 + y^2 + 4y + 4 + z^2 - 2z + 1 \]
So, the right side becomes:
\[ x^2 + y^2 + z^2 - 4x + 4y + 9 \]

### Step 5: Set the expanded forms equal
Now we equate the two expanded forms:
\[ x^2 + y^2 + z^2 - 4y - 6z + 13 = x^2 + y^2 + z^2 - 4x + 4y + 9 \]

### Step 6: Simplify the equation
Cancelling \(x^2\), \(y^2\), and \(z^2\) from both sides, we get:
\[ -4y - 6z + 13 = -4x + 4y + 9 \]
Rearranging gives:
\[ 4x - 8y - 6z + 4 = 0 \]

### Step 7: Divide by 4
Dividing the entire equation by 4 simplifies it to:
\[ x - 2y - \frac{3}{2}z + 1 = 0 \]

### Final Locus Equation
Thus, the locus of the point P which is equidistant from points A and B is given by:
\[ x - 2y - \frac{3}{2}z + 1 = 0 \]