I am very confused regarding a few problems in Calculus III. Any help or hints to any of these would be greatly appreciated!! 1.) Are lines L1 and L2 parallel? L1: (x-7)/6 = (y+5)/4 = -(z-9)/8; L2: -(x-11)/9 = -(y-7)/6 = (z-13)/12; The answer says that they are parallel, which I don`t understand. I know 2 lines are parallel if their direction vectors are parallel, but in this case V1 = <6, 4, -8>, and V2 = <9, -6, 12>. So they are not multiples of each other and thus I didn`t think they are parallel. What am I missing? 2.) Write an equation of the plane that contains both the point P and the line L: P(2,4,6); L: x = 7-3t, y = 3+4t, z = 5 + 2t; I know to write an equation of the plane you need a direction vector and a point. I tried using <-3,4,2> crossed with <2,4,6> as my normal vector and <2,4,6> as my (X0, Y0, Z0). But I got the wrong answer... 3.) Find an equation of the plane through P(3,3,1) that is perpendicular to the planes x+y = 2Z and 2X + z = 10. If I take the cross product of the second 2 planes that would give me a vector parallel to the equation that I want to find, but I need a normal vector. What to do? 4.) Find an equation of the plane that passes through the points P(1,0,-1), Q(2,1,0) and is parallel to the line of intersection of the planes x+y+z = 5 and 3x -y = 4. 5.) Prove that the lines x -1 = 1/2(y+1) = z-2 and x-2 = 1/3(y-2) = 1/2(z-4) intersect. Find an equation of the only plane that contains them both. Sorry for so many problems. But any help would be great! Thanks!