The equation of the plane passing through the point (1,1,1) and perpendicular to the planes 2x+y-z=5 and 3x-y-z=7, is

|ijk|2 1 -1|=i(-1+1)-j(-2+3)+k(-2-3)=-j+5k3 -1 -1Hence, -y+5z+c=0passes through (1,1,1) ∴ -1+5+c=0 ∴c=-4The equation of the plane is -y+5z-4=0 or y-5z+4=0.14x+2y+15z=31.So, option A is correct.