Question: A plane has $-4y + 6z - 4 = 0$ as its Cartesian equation. Determine the Cartesian equation of a plane that is perpendicular to and contain the point $P(-3, -10, 4)$.
I tried doing this question on my own but I messed up and I don't understand how I'm supposed to find the answer to this question. To solve the question, I tried to use the cross product but I got even more confused when doing it. Am I supposed to use the cross product? Or do I use another method? i would appreciate if anyone can help me out.
Take $y=0$ or $1$ and then you can get two position vector. Then the direction of the plane can be found by using cross product of the two direction you had obtained.