Solve this

Using the equations we get

$D=\left|\begin{array}{rrr}2 & 1 & -2 \\ 1 & -2 & 1 \\ 5 & -5 & 1\end{array}\right|$

$\Rightarrow 2(-2+5)-1(1-5)-2(-5+10)=0$

$D_{1}=\left|\begin{array}{ccc}4 & 1 & -2 \\ -2 & -2 & 1 \\ -2 & -5 & 1\end{array}\right|$

$\Rightarrow 4(-2+5)-1(-2+2)-2(10-4)=0$

$D_{2}=\left|\begin{array}{ccc}2 & 4 & -2 \\ 1 & -2 & 1 \\ 5 & -2 & 1\end{array}\right|$

$\Rightarrow 2(-2+2)-4(1-5)-2(-2+10)=0$

$D_{3}=\left|\begin{array}{ccc}2 & 1 & 4 \\ 1 & -2 & -2 \\ 5 & -5 & -2\end{array}\right|$

$\Rightarrow 2(4-10)-1(-2+10)+4(-5+10)=0$

$\therefore D=D_{1}=D_{2}=0$

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