If the system of equation `{:(,x-2y+z=a),(2x+y-2z=b),and,(x+3y-3z=c):}`
have at least one solution, then the relationalship between a,b,c is

Options

A. `a+b+c=0`
B. `a-b+c=0`
C. `-a+b+c=0`
D. `a+b-c=0`

Explanation

To find the relationship between \( a \), \( b \), and \( c \) for the given system of equations to have at least one solution, we can follow these steps:

### Step 1: Write down the given equations
The equations are:
1. \( x - 2y + z = a \) (Equation 1)
2. \( 2x + y - 2z = b \) (Equation 2)
3. \( x + 3y - 3z = c \) (Equation 3)

### Step 2: Set up the determinant
To determine the conditions for the system to have at least one solution, we need to calculate the determinant of the coefficients of the variables \( x, y, z \).

The coefficient matrix is:
\[ \begin{bmatrix} 1 & -2 & 1 \\ 2 & 1 & -2 \\ 1 & 3 & -3 \end{bmatrix} \]

### Step 3: Calculate the determinant
Let \( \Delta \) be the determinant of the coefficient matrix:
\[ \Delta = \begin{vmatrix} 1 & -2 & 1 \\ 2 & 1 & -2 \\ 1 & 3 & -3 \end{vmatrix} \]

Calculating this determinant using the formula for a 3x3 matrix:
\[ \Delta = 1 \cdot \begin{vmatrix} 1 & -2 \\ 3 & -3 \end{vmatrix} - (-2) \cdot \begin{vmatrix} 2 & -2 \\ 1 & -3 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} \]

Calculating the minors:
1. \( \begin{vmatrix} 1 & -2 \\ 3 & -3 \end{vmatrix} = (1)(-3) - (-2)(3) = -3 + 6 = 3 \)

2. \( \begin{vmatrix} 2 & -2 \\ 1 & -3 \end{vmatrix} = (2)(-3) - (-2)(1) = -6 + 2 = -4 \)

3. \( \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} = (2)(3) - (1)(1) = 6 - 1 = 5 \)

Now substituting back into the determinant:
\[ \Delta = 1 \cdot 3 + 2 \cdot 4 + 1 \cdot 5 = 3 + 8 + 5 = 16 \]

### Step 4: Set up the augmented matrix
Next, we set up the augmented matrix for the system:
\[ \begin{bmatrix} 1 & -2 & 1 & | & a \\ 2 & 1 & -2 & | & b \\ 1 & 3 & -3 & | & c \end{bmatrix} \]

### Step 5: Calculate the determinant of the augmented matrix
Let \( \Delta_a \) be the determinant of the augmented matrix:
\[ \Delta_a = \begin{vmatrix} 1 & -2 & 1 & a \\ 2 & 1 & -2 & b \\ 1 & 3 & -3 & c \end{vmatrix} \]

Using the same method as before, we replace the last column with \( a, b, c \) and calculate:
\[ \Delta_a = \begin{vmatrix} 1 & -2 & 1 \\ 2 & 1 & -2 \\ 1 & 3 & -3 \end{vmatrix} + a \cdot \begin{vmatrix} 1 & -2 \\ 1 & 3 \end{vmatrix} - b \cdot \begin{vmatrix} 1 & 1 \\ 1 & -3 \end{vmatrix} + c \cdot \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} \]

### Step 6: Solve for the relationship
For the system to have at least one solution, we need \( \Delta = 0 \) and \( \Delta_a = 0 \). This leads us to the relationship:
\[ a - b + c = 0 \]

### Final Answer
Thus, the relationship between \( a \), \( b \), and \( c \) is:
\[ a - b + c = 0 \]