Consider the following system of equations :
`{:(x+2y-3z=a),(2x+6y-11z=b),(x-2y+7z=c","):}`
where a, b and c are real constant. Then the system of eqations :

Options

A. has a unique solution when 5a=2b+c
B. has infinite number of solutions when 5a=2b+c
C. has no solution for all a, b and c
D. has a unique solutions for all a,b and c

Answer

To solve the given system of equations:

1. **Write down the equations**:
\[ \begin{align*} (1) & \quad x + 2y - 3z = a \\ (2) & \quad 2x + 6y - 11z = b \\ (3) & \quad x - 2y + 7z = c \end{align*} \]

2. **Form the coefficient matrix**:
The coefficient matrix \( A \) for the system is:
\[ A = \begin{bmatrix} 1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7 \end{bmatrix} \]

3. **Calculate the determinant of the coefficient matrix**:
We denote the determinant as \( \Delta \):
\[ \Delta = \begin{vmatrix} 1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7 \end{vmatrix} \]
We can calculate this determinant using the method of cofactor expansion.

Expanding along the first row:
\[ \Delta = 1 \cdot \begin{vmatrix} 6 & -11 \\ -2 & 7 \end{vmatrix} - 2 \cdot \begin{vmatrix} 2 & -11 \\ 1 & 7 \end{vmatrix} - 3 \cdot \begin{vmatrix} 2 & 6 \\ 1 & -2 \end{vmatrix} \]

Now calculating each of the 2x2 determinants:
\[ \begin{vmatrix} 6 & -11 \\ -2 & 7 \end{vmatrix} = (6 \cdot 7) - (-11 \cdot -2) = 42 - 22 = 20 \]
\[ \begin{vmatrix} 2 & -11 \\ 1 & 7 \end{vmatrix} = (2 \cdot 7) - (-11 \cdot 1) = 14 + 11 = 25 \]
\[ \begin{vmatrix} 2 & 6 \\ 1 & -2 \end{vmatrix} = (2 \cdot -2) - (6 \cdot 1) = -4 - 6 = -10 \]

Substituting back into the determinant:
\[ \Delta = 1 \cdot 20 - 2 \cdot 25 - 3 \cdot (-10) = 20 - 50 + 30 = 0 \]

4. **Determine the conditions for solutions**:
Since \( \Delta = 0 \), we need to check the conditions for infinite solutions. For infinite solutions, we need:
\[ \Delta = 0, \quad \Delta_x = 0, \quad \Delta_y = 0, \quad \Delta_z = 0 \]

5. **Calculate \( \Delta_x \)**:
Replace the first column of the coefficient matrix with the constants \( a, b, c \):
\[ \Delta_x = \begin{vmatrix} a & 2 & -3 \\ b & 6 & -11 \\ c & -2 & 7 \end{vmatrix} \]
Expanding this determinant similarly, we find:
\[ \Delta_x = 20a - 25b + 30c \]

6. **Set \( \Delta_x = 0 \)**:
For infinite solutions, we need:
\[ 20a - 25b + 30c = 0 \]

7. **Calculate \( \Delta_y \)**:
Replace the second column of the coefficient matrix with the constants \( a, b, c \):
\[ \Delta_y = \begin{vmatrix} 1 & a & -3 \\ 2 & b & -11 \\ 1 & c & 7 \end{vmatrix} \]
Similarly, we find:
\[ \Delta_y = 20b - 25a + 30c = 0 \]

8. **Calculate \( \Delta_z \)**:
Replace the third column of the coefficient matrix with the constants \( a, b, c \):
\[ \Delta_z = \begin{vmatrix} 1 & 2 & a \\ 2 & 6 & b \\ 1 & -2 & c \end{vmatrix} \]
Similarly, we find:
\[ \Delta_z = 20c - 25a + 30b = 0 \]

9. **Final condition for infinite solutions**:
The condition for infinite solutions can be summarized as:
\[ 5a = 2b + c \]

Thus, the system of equations has infinite solutions when \( 5a = 2b + c \).