I am helping my brother. It's been awhile since I have done linear algebra or Calculus III, but I remember this sounds like ODE's $y=y_c+y_p$.
Problem:
Suppose $\vec a=[1 \ 0 \ 1]^T, \vec b=[2 \ 7 \ -2]^T, \vec c=[3 \ 1\ 5]^T$ are lying on a plane in the 3D space. Prove that $\vec d=[4 \ 8 \ 2]^T$ also lies on that plane.
Hint:
A plane can be expressed as a particular solution plus the span of linearly independent vectors.
Considering that the 3 given vectors are linearly independent and thus span $\mathbb R^3$, what is a plane that they lie on?
What does the hint mean here?
My brother was not taught cross product yet.
Best Answer
My brother got clarification:
The vectors do not lie on a plane. The terminal points of the vectors lie on a plane.
mr_e_man and Chris Custer are right.