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

We have ,x+ky+3z=0;  3x+ky−2z=0;  2x+4y−3z=0System of equation has non-zero solution, if 1k33k−224−3=0⇒(−3k+8)−k(−9+4)+3(12−2k)=0⇒ −3k+8+9k−4k+36−6k=0⇒−4k+44=0⇒k=11 Let  z=λ, then we get x+11y+3λ=0----13x+11y−2λ=0-----2 And  2x+4y−3λ=0----3Solving Eqs. (i) and (ii), we get x=5λ2,y=−λ2,z=λ⇒xzy2=5λ22×−λ22=10