If the system of linear equations x+ky+3z=0, 3x+ky-6z=0, 2x+4y-7z=0 has a non-zero solution (x,y,z), then xzy2 is equal to

x+ky+3x=0.. (i)
3x+ky-6z=0.. (ii)
2x+4y-7z=0.. (iii)
(i)-(ii)→3x+x+ky-ky-6z+7z=0
2x=5z
-(4)
(i)+(iii)→x+2x+ky+4y=0
3x=-(k+4)y
3y/2=x (25)
x(-7k+4)/2.x2=2(k+4)/25×9=2(k+4)/45
Also, for non-zero soln to exist,|A|=0
→∣∣∣∣1k33k-624-7∣∣∣∣=0
-7k+8-k( -18+4)+3(12-6k)=0
-7k+8+14k+36-18k=0
-11k+44=0
k=11
∴x/3y/2=2/45(k+4)/2=2×15×15/45
x/3y/2=10