To solve the given system of equations using Gaussian elimination, we start by writing the augmented matrix for the system:
[tex]\[
\begin{pmatrix}
5 & 12 & 7 & | & 11 \\
2 & 5 & 5 & | & -2 \\
1 & 2 & -3 & | & 6
\end{pmatrix}
\][/tex]
We will perform Gaussian elimination to try to reach the reduced row echelon form (RREF).
### Step 1: Form Row Echelon Form
1. First pivot: Use the (1,1) position as the first pivot to eliminate the entries in the first column below it.
[tex]\[
R2 = R2 - \left(\frac{2}{5}\right) R1 \quad \text{and} \quad R3 = R3 - \left(\frac{1}{5}\right) R1
\][/tex]
This gives:
[tex]\[
\begin{pmatrix}
5 & 12 & 7 & | & 11 \\
0 & \left(5 - \frac{24}{5}\right) & \left(5 - \frac{14}{5}\right) & | & \left(-2 - \frac{22}{5}\right) \\
0 & \left(2 - \frac{12}{5}\right) & \left(-3 - \frac{7}{5}\right) & | & \left(6 - \frac{11}{5}\right)
\end{pmatrix}
\][/tex]
Simplify the rows:
[tex]\[
R2: \quad 0, \quad \frac{1}{5}, \quad \frac{11}{5}, \quad \left| \quad \frac{-32}{5} \quad \right.
\][/tex]
[tex]\[
R3: \quad 0, \quad \frac{-2}{5}, \quad - \frac{22}{5}, \quad \left| \quad \frac{19}{5} \quad \right.
\][/tex]
Simplified matrix:
[tex]\[
\begin{pmatrix}
5 & 12 & 7 & | & 11 \\
0 & \left(\frac{1}{5} \cdot 5\right) & \left(\frac{11}{5}\cdot 5\right) & | & \left(\frac{-32}{5}\cdot 5\right) \\
0 & \left(\frac{-2}{5} \cdot 5\right) & \left(\frac{-22}{5} \cdot 5\right) & | & \left(\frac{19}{5} \cdot 5\right)
\end{pmatrix}
\][/tex]
[tex]\[
\begin{pmatrix}
5 & 12 & 7 & | & 11 \\
0 & 1 & 11 & | & -6 \\
0 & -2 & -22 & | & 19
\end{pmatrix}
\][/tex]
### Step 2: Eliminate below the second row's pivot.
2. Second pivot: Use the (2,2) position as the pivot to eliminate the entries in the second column below it.
[tex]\[
R3 = R3 + 2R2
\][/tex]
Simplified matrix:
[tex]\[
\begin{pmatrix}
5 & 12 & 7 & | & 11 \\
0 & 1 & 11 & | & -6 \\
0 & 0 & 0 & | & 7
\end{pmatrix}
\][/tex]
### Step 3: Interpretation.
Notice that after reaching the Row Echelon Form, our system has a resulting matrix row:
[tex]\[
0 \quad 0 \quad 0 \quad \mid \quad 7
\][/tex]
Such a row indicates inconsistency. It translates to the equation:
[tex]\[
0 = 7
\][/tex]
This contradicts any possible solution because such an equation is never true.
### Conclusion
Given this inconsistency, we conclude that the system of equations has no solution. Thus, we select:
[tex]\[
\boxed{\text{D. There is no solution.}}
\][/tex]
