The system of equations
`(a-b) x+(b-c) y +(c - a) z=0`
`(b-c) x+(c - a) y +(a-b) z=0`
`(c- a) x+(a-b) y +(b-c) z=1`
has no solution.

Answer

To determine whether the given system of equations has no solution, we can analyze the system using determinants. The equations are:

1. \((a-b)x + (b-c)y + (c-a)z = 0\)
2. \((b-c)x + (c-a)y + (a-b)z = 0\)
3. \((c-a)x + (a-b)y + (b-c)z = 1\)

We can express this system in matrix form as \(A\mathbf{x} = \mathbf{b}\), where:

\[ A = \begin{pmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \]

### Step 1: Calculate the Determinant of Matrix A

To check for the existence of solutions, we first need to calculate the determinant of matrix \(A\). If \(\det(A) = 0\), the system may have either no solution or infinitely many solutions.

\[ \det(A) = \begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix} \]

### Step 2: Use Row Operations to Simplify the Determinant

We can simplify the determinant by performing row operations. Let's add all three rows together:

\[ R_1 + R_2 + R_3 \rightarrow R_1 \]

This gives us:

\[ \begin{pmatrix} 0 & 0 & 0 \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{pmatrix} \]

### Step 3: Analyze the Resulting Matrix

Since the first row of the determinant is now all zeros, we can conclude that:

\[ \det(A) = 0 \]

### Step 4: Check for Consistency

Since \(\det(A) = 0\), we need to check the consistency of the system. We will calculate the determinants for the modified matrices for \(x\), \(y\), and \(z\) (using Cramer's rule).

#### Determinant for \(x\):

Replace the first column of \(A\) with \(\mathbf{b}\):

\[ \det(A_x) = \begin{vmatrix} 0 & b-c & c-a \\ 0 & c-a & a-b \\ 1 & a-b & b-c \end{vmatrix} \]

Expanding this determinant will yield a non-zero value, indicating that the system is inconsistent.

### Conclusion

Since \(\det(A) = 0\) and \(\det(A_x) \neq 0\), the system of equations has no solution.