Solve this

Using the equations we get

$D=\left|\begin{array}{rrr}1 & -1 & 1 \\ 2 & 1 & -1 \\ -1 & -2 & 2\end{array}\right|$

$\Rightarrow 1(2-2)+1(4-1)+1(-4+1)=0$

$D_{1}=\left|\begin{array}{rrr}3 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & 2\end{array}\right|$

$\Rightarrow 3(2-2)+1(4+1)+1(-4-1)=0$

$D_{2}=\left|\begin{array}{rrr}1 & 3 & 1 \\ 2 & 2 & -1 \\ -1 & 1 & 2\end{array}\right|$

$\Rightarrow 1(4+1)-3(4-1)+1(2+2)=0$

$D_{3}=\left|\begin{array}{rrr}1 & -1 & 3 \\ 2 & 1 & 2 \\ -1 & -2 & 1\end{array}\right|$

$\Rightarrow 1(1+4)+1(2+2)+3(-4+1)=0$

Here,

$D=D_{1}=D_{2}=D_{3}=0$

Thus, the system of linear equations has infinitely many solutions.