The augmented matrix $A|B$ would like this: $\begin{bmatrix} 1& 2& 1& 6\\ 2& 1& 2& 6\\ 1& 1& 1& 5 \end{bmatrix}$
We will find the rank of $A$ i.e. coefficient matrix and also rank of $A|B$ i.e. augmented matrix
$R_{1}\leftrightarrow R_{3}$ $\Rightarrow$ $\begin{bmatrix} 1& 1& 1& 5\\ 2& 1& 2& 6\\ 1& 2& 1& 6 \end{bmatrix}$
$R_{2}\rightarrow R_{2}-R_{1}$ and $R_{3}\rightarrow R_{3}-R_{1}$ $\Rightarrow$ $\begin{bmatrix} 1& 1& 1& 5\\ 0& -1& 0& -4\\ 0& 1& 0& 1 \end{bmatrix}$
$R_{3}\rightarrow R_{3}+R_{2}$ $\Rightarrow$ $\begin{bmatrix} 1& 1& 1& 5\\ 0& -1& 0& -4\\ 0& 0& 0& -3 \end{bmatrix}$
Now, $\rho (A)=3-1=2$ and $\rho(A|B)=3$
Since, $\rho (A) < \rho(A|B)$, the system of equations is inconsistent and hence has no solution.
Option C is correct.