Step 1 :We start by writing the system of equations in augmented matrix form: \[\begin{bmatrix} 5 & 12 & 5 & 13 \\ 2 & 5 & 5 & -8 \\ 1 & 2 & -5 & 3 \end{bmatrix}\]
Step 2 :We then perform row operations to get the matrix in row-echelon form. First, we swap the first and third rows to get: \[\begin{bmatrix} 1 & 2 & -5 & 3 \\ 2 & 5 & 5 & -8 \\ 5 & 12 & 5 & 13 \end{bmatrix}\]
Step 3 :Next, we subtract twice the first row from the second row and five times the first row from the third row to get: \[\begin{bmatrix} 1 & 2 & -5 & 3 \\ 0 & 1 & 15 & -14 \\ 0 & 2 & 30 & -2 \end{bmatrix}\]
Step 4 :We then subtract twice the second row from the third row to get: \[\begin{bmatrix} 1 & 2 & -5 & 3 \\ 0 & 1 & 15 & -14 \\ 0 & 0 & 0 & 30 \end{bmatrix}\]
Step 5 :From the third row, we see that 0 = 30, which is a contradiction. Therefore, the system of equations has no solution.
Step 6 :So, the correct choice is C. There is no solution. The solution set is \(\varnothing\).