To find the point on the ZX plane that is equidistant from the points \( A(1, -1, 0) \), \( B(2, 1, 2) \), and \( C(3, 2, -1) \), we can follow these steps:
### Step 1: Define the point on the ZX plane
Since the point lies on the ZX plane, we can represent it as \( P(x, 0, z) \), where the y-coordinate is 0.
### Step 2: Set up the distance equations
The point \( P \) must be equidistant from points \( A \), \( B \), and \( C \). Therefore, we can set up the following equations based on the distance formula:
1. Distance from \( P \) to \( A \):
\[
PA = \sqrt{(x - 1)^2 + (0 + 1)^2 + (z - 0)^2}
\]
2. Distance from \( P \) to \( B \):
\[
PB = \sqrt{(x - 2)^2 + (0 - 1)^2 + (z - 2)^2}
\]
3. Distance from \( P \) to \( C \):
\[
PC = \sqrt{(x - 3)^2 + (0 - 2)^2 + (z + 1)^2}
\]
### Step 3: Equate distances
To find \( P \), we can first equate the distances \( PA \) and \( PB \):
\[
(x - 1)^2 + (0 + 1)^2 + (z - 0)^2 = (x - 2)^2 + (0 - 1)^2 + (z - 2)^2
\]
### Step 4: Expand and simplify the equation
Expanding both sides:
\[
(x - 1)^2 + 1 + z^2 = (x - 2)^2 + 1 + (z - 2)^2
\]
This simplifies to:
\[
(x - 1)^2 + z^2 = (x - 2)^2 + (z^2 - 4z + 4)
\]
Expanding further:
\[
x^2 - 2x + 1 + z^2 = x^2 - 4x + 4 + z^2 - 4z + 4
\]
Cancelling \( z^2 \) and \( x^2 \):
\[
-2x + 1 = -4x + 8 - 4z
\]
Rearranging gives:
\[
2x - 4z = 7 \quad \text{(Equation 1)}
\]
### Step 5: Equate distances \( PB \) and \( PC \)
Now, equate \( PB \) and \( PC \):
\[
(x - 2)^2 + (0 - 1)^2 + (z - 2)^2 = (x - 3)^2 + (0 - 2)^2 + (z + 1)^2
\]
### Step 6: Expand and simplify this equation
Expanding both sides:
\[
(x - 2)^2 + 1 + (z - 2)^2 = (x - 3)^2 + 4 + (z + 1)^2
\]
This simplifies to:
\[
(x - 2)^2 + (z - 2)^2 = (x - 3)^2 + (z + 1)^2 + 3
\]
Expanding further:
\[
x^2 - 4x + 4 + z^2 - 4z + 4 = x^2 - 6x + 9 + z^2 + 2z + 1 + 3
\]
Cancelling \( z^2 \) and \( x^2 \):
\[
-4x + 8 = -6x + 13 + 2z
\]
Rearranging gives:
\[
2x - 2z = 5 \quad \text{(Equation 2)}
\]
### Step 7: Solve the system of equations
Now we have the system of equations:
1. \( 2x - 4z = 7 \)
2. \( 2x - 2z = 5 \)
Subtract Equation 2 from Equation 1:
\[
(2x - 4z) - (2x - 2z) = 7 - 5
\]
This simplifies to:
\[
-2z = 2 \implies z = -1
\]
### Step 8: Substitute \( z \) back to find \( x \)
Substituting \( z = -1 \) into Equation 2:
\[
2x - 2(-1) = 5 \implies 2x + 2 = 5 \implies 2x = 3 \implies x = \frac{3}{2}
\]
### Final Answer
Thus, the point \( P \) on the ZX plane that is equidistant from points \( A \), \( B \), and \( C \) is:
\[
P\left(\frac{3}{2}, 0, -1\right)
\]
