To determine the value of \(\lambda\) such that the given system of equations has an infinite number of solutions, we need to consider the criterions for a system of equations to have infinite solutions. For a system of linear equations to have infinite solutions, the equations must be dependent, i.e., the coefficient matrix should have a rank less than the number of variables, or equivalently the augmented matrix should have the same rank as the coefficient matrix.
The given system of equations is:
\[ \begin{align*} x - 3y + 5z &= 3, \\ x - 2y + 4z &= 4, \\ 2x - 7y + \lambda z &= 5. \end{align*} \]
Let's set up the coefficient matrix \(A\) and the augmented matrix [A|B]:
| Coefficient Matrix \(A\): | \[ \begin{bmatrix} 1 & -3 & 5 \\ 1 & -2 & 4 \\ 2 & -7 & \lambda \end{bmatrix} \] |
| Augmented Matrix [A|B]: | \[ \begin{bmatrix} 1 & -3 & 5 & | & 3 \\ 1 & -2 & 4 & | & 4 \\ 2 & -7 & \lambda & | & 5 \end{bmatrix} \] |
Subtract the first row from the second:
\[ R_2 \rightarrow R_2 - R_1 \quad \Rightarrow \quad \begin{bmatrix} 1 & -3 & 5 & | & 3 \\ 0 & 1 & -1 & | & 1 \\ 2 & -7 & \lambda & | & 5 \end{bmatrix} \]
Subtract 2 times the first row from the third row:
\[ R_3 \rightarrow R_3 - 2R_1 \quad \Rightarrow \quad \begin{bmatrix} 1 & -3 & 5 & | & 3 \\ 0 & 1 & -1 & | & 1 \\ 0 & -1 & \lambda-10 & | & -1 \end{bmatrix} \]
Now, add Row 2 to Row 3:
\[ R_3 \rightarrow R_3 + R_2 \quad \Rightarrow \quad \begin{bmatrix} 1 & -3 & 5 & | & 3 \\ 0 & 1 & -1 & | & 1 \\ 0 & 0 & \lambda-11 & | & 0 \end{bmatrix} \]
For the system to have infinitely many solutions, the rank of the coefficient matrix \(A\) must equal the rank of the augmented matrix [A|B], and both must be less than 3. Therefore, we require:
\[ \lambda - 11 = 0 \quad \Rightarrow \quad \lambda = 11. \]
Thus, the value of \(\lambda\) is 11. Therefore, the correct answer is \(\lambda = 11\).