What is the ordered triples where the plane cuts the y-axis for this equation 3x plus 4y plus 6z equals 12?

To find the ordered triple where the plane defined by the equation (3x + 4y + 6z = 12) cuts the y-axis, we set (x = 0) and (z = 0). Substituting these values into the equation gives (4y = 12), resulting in (y = 3). Therefore, the ordered triple where the plane intersects the y-axis is ((0, 3, 0)).