So just as the professor told you, you found the equation of the plane which passes through the origin. However, that is only the case when you are either told that, or your points happen to be given such that the plane defined by the three points contains the origin. If it helps, imagine that by assuming your plane went through the origin, you essentially created a parallel plane to the plane defined through the three points, and your parallel plane was created to pass through the origin, but have the same 'tilt' as the plane which passes through the three points.
Explicitly, you should have done something like: form the vectors $(1,0, 0)$ from the first two, and $1, 1, 0$ from the second two. Then their cross product is $(0,0,1)$ to give you a normal vector. Using this normal vector, we can use a point on the line and a normal vector to write the equation of a plane as: our normal vector: $(n1, n2, n3) = (0, 0, 1)$, and our point (we could use any of them, lets just choose the first) $(p1, p2, p3) = (0,1,2)$. Then our plane will be $n1(x-p1) + n2(y-p2) + n3(z-p3)$ which in this case is $0 + 0 + 1(z-2) =0 $ or $ z=2$. You can check to make sure that this works for our three points by just seeing that all of their third coordinates are in fact $2$. Hope that answers what you were still uncertain about.
It’s true that any set of three points is coplanar, but for linear dependence, we only care about planes that pass through the origin, as those are the two-dimensional vector subspaces of $\mathbb R^3$. Those planes are the set of all linear combinations of a pair of linearly-independent vectors, whereas planes that don’t include the origin can’t be generated that way. So, in the context of linear algebra, the qualifier “through the origin” is often understood and omitted to save space.
Best Answer
Hint
These vectors are coplanar if and only if $$(v_1\wedge v_2)\wedge v_3=0$$