I think I found it. I would need to take the cross product of 2 vectors from those 3 points to find the normal vector. Then I would use the equation 0 = a(x-x1)+b(y-y1)+c(z-z1), <a,b,c> being the normal vector. For <x1,y1,z1> would I just pick one of the points? Does it matter which one?
That form of the plane equation works but I prefer to write the plane equation as ax+by+cz=d where <a,b,c> is a normal vector to the plane. You can then solve for d by evaluating the left-hand side at any point. In the end it doesn't matter because you'll end up with the same equation.