Find the locus of a point whose each point is equidistant from the points A(2,3,-4) and B(-1,2,3).

Explanation

To find the locus of a point that is equidistant from the points A(2, 3, -4) and B(-1, 2, 3), we can follow these steps:

### Step 1: Define the points and the point P
Let the point P be represented as \( P(x, y, z) \). The given points are:
- \( A(2, 3, -4) \)
- \( B(-1, 2, 3) \)

### Step 2: Set up the distance formula
The condition for point P to be equidistant from points A and B can be expressed as:
\[ PA = PB \]

Using the distance formula in three dimensions, we have:
\[ PA^2 = PB^2 \]

### Step 3: Calculate \( PA^2 \) and \( PB^2 \)
Using the distance formula:
\[ PA^2 = (x - 2)^2 + (y - 3)^2 + (z + 4)^2 \]
\[ PB^2 = (x + 1)^2 + (y - 2)^2 + (z - 3)^2 \]

### Step 4: Set the equations equal to each other
Setting \( PA^2 \) equal to \( PB^2 \):
\[ (x - 2)^2 + (y - 3)^2 + (z + 4)^2 = (x + 1)^2 + (y - 2)^2 + (z - 3)^2 \]

### Step 5: Expand both sides
Expanding the left side:
\[ (x - 2)^2 = x^2 - 4x + 4 \]
\[ (y - 3)^2 = y^2 - 6y + 9 \]
\[ (z + 4)^2 = z^2 + 8z + 16 \]
So,
\[ PA^2 = x^2 - 4x + 4 + y^2 - 6y + 9 + z^2 + 8z + 16 \]
Combining these gives:
\[ PA^2 = x^2 + y^2 + z^2 - 4x - 6y + 8z + 29 \]

Now expanding the right side:
\[ (x + 1)^2 = x^2 + 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
\[ (z - 3)^2 = z^2 - 6z + 9 \]
So,
\[ PB^2 = x^2 + 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 \]
Combining these gives:
\[ PB^2 = x^2 + y^2 + z^2 + 2x - 4y - 6z + 14 \]

### Step 6: Set the expanded forms equal
Now we set the two expanded forms equal to each other:
\[ x^2 + y^2 + z^2 - 4x - 6y + 8z + 29 = x^2 + y^2 + z^2 + 2x - 4y - 6z + 14 \]

### Step 7: Simplify the equation
Cancelling \( x^2, y^2, z^2 \) from both sides:
\[ -4x - 6y + 8z + 29 = 2x - 4y - 6z + 14 \]

Rearranging gives:
\[ -4x - 2x + 4y - 6y + 8z + 6z + 29 - 14 = 0 \]
This simplifies to:
\[ -6x - 2y + 14z + 15 = 0 \]

### Step 8: Rearranging the equation
Rearranging gives:
\[ 6x + 2y - 14z = 15 \]

### Conclusion
The locus of the point \( P \) that is equidistant from points \( A \) and \( B \) is given by the equation:
\[ 6x + 2y - 14z = 15 \]