vectors list:

A(2,-1,-2)

B(1,2,1)

C(2,3,0)

D(5,0,-6)

if $(\stackrel{\rightharpoonup}{AB}\times \stackrel{\rightharpoonup}{AC})\xb7\stackrel{\rightharpoonup}{AD}=0$ then all vectors are on the same plane.

$\stackrel{\rightharpoonup}{AB}=(-1,3,3)=(1-2,2-(-1),1-(-2\left)\right)$

$\stackrel{\rightharpoonup}{AC}=(0,4,2)$

$\stackrel{\rightharpoonup}{AD}=(3,1,-4)$

$\left|\begin{array}{ccc}-1& 3& 3\\ 0& 4& 2\\ 3& 1& -4\end{array}\right|=36-36=0$

vectors are on the same plane.