1.The number of solutions (x,y,z)(x, y, z)(x,y,z) to the system of equations x+2y+4z=9x + 2y + 4z = 9x+2y+4z=9, 4yz+2xz+xy=134yz + 2xz + xy = 134yz+2xz+xy=13, xyz=3xyz = 3xyz=3 such that at least two of x,y,zx, y, zx,y,z are integers isa.333b.555c.666d.444Login to continueOnly logged in users canattempt or see the solution.
