If the system of linear equations x+ky+3z=0 ,3x+ky−2z=0,2x+4y−3z=0has a non – zero solutionx,y,zthen xzy2=

The system has a non-zero solution⇒Δ=0 ⇒1k33k−224−3=0 ⇒1-3k+8-k-9+4+312-2k=0 ⇒4k=44 ⇒k=11 ∴The given equations become x+11y+3z=0, 3x+11y-2z=0 and 2x+4y-3z=0 Solving the above, we get x-22-33=y9+2=z22-33=rsay ⇒x=-5r, y=r, z=-2r ∴xzy2=-5r-2rr2=10