to get the equation of plane ABC you need the normal vector n of the plane which is orthogonal with the plane. It is AB cross BC . As previous part we have, vector n = <-12,4,4> . so a = -12, b= 4, c = 4
then apply to formula for equation of the plane . I choose the point A(1,2,-1) to apply. that is a(x-1) + b (y-2) +c(z+1)=0 . At the end up, the equation is: _12(x-1) +4(y-2) +4 (z-1)=0 or you can expand it to get more simple form.