If lines 2x−3y+6=0 and  kx+2y+12=0 cut the coordinate axes in the concylic points, then the value of k is

2x−3y+6=0 meets the axes at A−3,0 and B0,2. kx+2y+12=0 meets the axes at the D0,−6 and C−12/k,0.OA×OC=OB×OD⇒3×12/k=2×6⇒k=−3