第2045题:linear equations
If the system of linear equations
{x+ky+3z=03x+ky−2z=02x+4y−3z=0\begin{cases} x+ky+3z=0 \\ 3x+ky-2z=0 \\ 2x+4y-3z=0 \end{cases}⎩⎪⎨⎪⎧x+ky+3z=03x+ky−2z=02x+4y−3z=0
has a non-zero slution (x,y,z) (x,y,z)(x,y,z) , then xzy2\dfrac{xz}{y^2}y2xz is ( ).