I am given 3 vectors. $u=[1, 2, 3], v=[-3, 1, -2], w=[2,-3,-1]$. And the question is how do I know that $u, v, w$ lie in a plane? I can't understand what they mean. Because for me each of them fills some line. And only their pairwise linear combinations can fill a plane, and all triple linear combinations fill 3d space. So how should I answer the question? Thanks in advance.
UPD1: This is from book "Introduction to Linear Algebra - 4th edition", Gilbert Strang. Page 6, problem 5.
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Three vectors $u,v,w$ always lie in the plane parallel to the subspace spanned by $v-u$ and $w-u$ that passes through $u$. It is the set of all combinations $$u+a(v-u)+b(w-u)$$ or equivalently $$au+bv+cw$$ where $a+b+c=1$.
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One way is to show the cross product of one vector with the other two yields two vectors which are parallel, ie. There unit vectors are equal or negative of the other.
You could even define a plane in $\Bbb R^3$ to be a set of vectors,
$ P = \{\mathbf x \in \Bbb R^3 \ | \ \mathbf a , \mathbf b \in P \ \Rightarrow \ \frac{\mathbf x \times \mathbf a}{|| \mathbf x \times \mathbf a ||} = \frac{\mathbf x \times \mathbf b}{|| \mathbf x \times \mathbf b ||} \lor \frac{\mathbf x \times \mathbf a}{|| \mathbf x \times \mathbf a ||} = -1\frac{\mathbf x \times \mathbf b}{|| \mathbf x \times \mathbf b ||} \}$.
Then three vectors $u,v,w$ are coplanar iff $u,v,w \in P$. That is they are contained in some plane $P$ defined above.
The question is whether or not these vectors are linearly independent. A nice practical criterion is a determinant of a matrix cinsisting of these vectors. They are linearly independent iff this determinant is non-zero. This is not the case here, so $u,v,w$ are complanar (i.e. linearly dependent) vectors. Try to express $u$ as a linear combination of $v,w$.