Question:
$x+y-z=0$
$x-2 y+z=0$
$3 x+6 y-5 z=0$
Solution:
Using the equations we get
$D=\left|\begin{array}{ccr}1 & 1 & -1 \\ 1 & -2 & 1 \\ 3 & 6 & -5\end{array}\right|$
$\Rightarrow 1(10-6)-1(-5-3)-1(6+6)=0$
$D_{1}=\left|\begin{array}{ccr}0 & 1 & -1 \\ 0 & -2 & 1 \\ 0 & 6 & -5\end{array}\right|$
$\Rightarrow 0(10-6)-1(0-0)-1(0+0)=0$
$D_{2}=\left|\begin{array}{rrr}1 & 0 & -1 \\ 1 & 0 & 1 \\ 3 & 0 & -5\end{array}\right|$
$\Rightarrow 1(0-0)-0(-5-3)-1(0-0)=0$
$D_{3}=\left|\begin{array}{ccc}1 & 1 & 0 \\ 1 & -2 & 0 \\ 3 & 6 & 0\end{array}\right|$
$\Rightarrow 1(0-0)-1(0-0)+0(6+6)=0$
$\therefore D=D_{1}=D_{2}$
Hence, the system of linear equations has infinitely many solutions.