is this right?

Compute the vector cross product of <2, -3, 4> and <-1, 2, 4>. This gives you a vector that is perpendicular to both given planes. <-20, -12, 1>

Let Q(x, y, z) be a point in the sought plane. Then vector PQ is perpendicular to <-20, -12, 1>, and so their dot product is zero:

<-20, -12, 1> • <x-1, y-2, z-0> = 0

-20(x-1) - 12(y-2) + 1(z-0) = 0

20(x-1) + 12(y-2) - z = 0

20x + 12y - z = 44