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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.
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