Use Gaussian elimination to find the complete solution to the following system of equations, or show that none exists.
$$\{\begin{array}{rr}5x+12y+5z=& 13\\ 2x+5y+5z=& -8\\ x+2y-5z=& 3\end{array}$$
Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
A. There is one solution. The solution set is $\{(,,D)\}$. (Simplify your answers.)
B. There are infinitely many solutions. The solution set is $\{(,;,z)\}$, where $z$ is any real number. (Type expressions using $z$ as the variable. Use integers or fractions for any numbers in the expression.)
C. There is no solution. The solution set is $\varnothing$.

Step 1 ：We start by writing the system of equations in augmented matrix form: $$\left[\begin{array}{cccc}5& 12& 5& 13\\ 2& 5& 5& -8\\ 1& 2& -5& 3\end{array}\right]$$

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: $$\left[\begin{array}{cccc}1& 2& -5& 3\\ 2& 5& 5& -8\\ 5& 12& 5& 13\end{array}\right]$$

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: $$\left[\begin{array}{cccc}1& 2& -5& 3\\ 0& 1& 15& -14\\ 0& 2& 30& -2\end{array}\right]$$

Step 4 ：We then subtract twice the second row from the third row to get: $$\left[\begin{array}{cccc}1& 2& -5& 3\\ 0& 1& 15& -14\\ 0& 0& 0& 30\end{array}\right]$$

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 $\varnothing$.