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**ANDS!** The point P(1,2,1) is not on the line of intersection of those two planes. It certainly works in the first equation, but it does not satisfy the second equation, thus that point is not on plane two, and it can not be in the line of intersection of the two points. It is a simple error that you have done, probably smack your head when you see it.

This is correct, although you do not need to get the vector equation, only the direction vector <2, 13, 7>. In part a), the want the plane to pass through the point A(3,1,3). Taking the point in the line of intersection that crosses the x-plane, Q(0,-7/2, -3/2), you can use point A and point Q to get the vector AQ (or QA). Now you have two vectors (and numerous points). You should be able here to find the normal vector to the plane (since you have two vectors), and use one of the points to define your plane.

Part B is correct.

Way too much work. If a plane is parallel to another plane, then they the normal vector of the first plane is also a normal to the second plane. Thus, using the same point from part a), and the normal, you can solve B in seconds, without actually taking any cross-products and what-nots.