Well, From what I understand, to determine the intersection of a line and a plane, we use parametric form of the line and substitute the values of x, y and z into the Cartesian equation of the plane, correct? so, given the line x = 2 + 4t y = -1 + kt <=== note the 'k' variable z = 5 - 3t and the plane 7x + 6y - z - 3 = 0 What must be the value of k for no intersection point, one intersection point, an infinite number of intersection points? *** My calculations this far are written below. So far I think I'm headed completely in the wrong direction but I've exhausted the only way I thought possible. If I plug the parametric form into the plane equation, I end up with 31t + 6kt = 0 this is where I'm lost. The question has me lost, because I have not tried this form (usually I have been given a non-variable value for the parametric equations, plugging them in is easy and I go from there) At this point, I'm -guessing- that to have no intersection point, I must have a constant on the other side of the equation and be inconsistent (of the sort, 0t = 123). This is not possible as I can't just invent one? So, there cannot be NO intersection (it must pass through the plane) To have one intersection point, we can suppose any value of k, then t = 0? And for infinite number of intersection points, basically I need 0t = 0 (dependent system) SO, by letting k = - 31/6 31t + 6kt = 0 0t = 0 and thus an infinite number of intersection points (the line is contained in the plane) *** The above is probably incorrect ( I feel) but I do not know another way. Please suggest the correct way of going about this. Do not give me the answer, I would rather just some advice on how to work my way back on this question!