If the system of equations 2x+3ky+(3k+4)z=0, x+(k+4)y+(4k+2)z=0, x+2(k+4)y+(3k+4)z=0 has a non-trivial solution, then k=

The system has non-trivial solution⇒Δ=0⇒23k3k+41k+44k+212(k+4)3k+4=0, Apply R2→2R2-R1, R3→2R3-R1⇒23k3k+40-k+85k0k+163k+4=0⇒2[(-k+8)(3k+4)-5k(k+16)]=0⇒-8k2-60k+32=02k2+15k-8=0∴k=-8 or 12.