Determining the span of the vector space (linear Algebra) Find a vector $v$ that is not in  span{$(1, 1, 1),(1,2,3)$}
What i tried 
From the definition of the span, i know that the vector $v$ must be a vector that cannot be  expressed as a linear combination of the vectors $(1, 1, 1)$ and $(1,2,3)$
I choose the vector $v$ to be $(0,0,0)$. Im unsure whether this is correct
The point is that $(0,0,0)=0(1, 1, 1)+0(1,2,3)$ and so there still is a trival solution.(I also checked that there are no other possible linear combinations) And so at first glance this seems to be a linear combination. But $0$ is not considered a scalar so that is not considered as a linear combination. And thus $(0,0,0)$ is not in the span Am i correct?
Also i tried to use another argument to show that the vector $v=(0,0,0)$ is indeed not in the span.
I let $k_{1}$ and $k_{2}$ represent the vectors $(1, 1, 1)$ and $(1,2,3)$ respectively and i assume that the vector $v=(0,0,0)$ lies in the span. 
$v$ can be expressed as a linear combination of $k_{1}$ and $k_{2}$, hence we have $v=c_{1}k_{1}+c_{2}k_{2}$ then from the definition of the span we have a vector $w$ that can be written as a linear combination in the following form
$$w=a_{1}k_{1}+a_{2}k_{2}+a_{3}v$$
substituting in $v=c_{1}k_{1}+c_{2}k_{2}$
we have the following expression 
$$w=a_{1}k_{1}+a_{2}k_{2}+a_{3}(c_{1}k_{1}+c_{2}k_{2})$$
Simplifying the expression, we get 
$$w=(a_{1}+a_{3}c_{1})k_{1}+(a_{2}+a_{3}c_{2})k_{2}$$
But since $c_{1}=0$ and  $c_{2}=0$
then $$w=a_{1}k_{1}+a_{2}k_{2}$$ which is just simply the span of the first two vectors. Am i correct? Could anyone explain. Thanks
 A: $(0,0,0)$ is in the span, unfortunately.
A simple one not in the span would be $(0,1,1)$ as $A(1,1,1)+B(1,2,3)=(0,1,1)$ has no solution. The second and third term means $B=0$ but $A$ cannot be anything to generate $(0,1,1)$.
For your argument, let's sonsider the following simple example:
Does $(0,1)$ lies in the span by $(0,1)$ and $(1,0)$?
$(0,1)=1(0,1)+0(1,0)$ is the only solution and you cannot deny using $0$ as a scalar otherwise $(0,1)$ won't be in its own span!
A: What you want is to find $a,b,c$ such that the linear system
$$
x(1,1,1)+y(1,2,3)=(a,b,c)
$$
has no solution. Let's do Gaussian elimination on the matrix:
\begin{align}
\begin{bmatrix}
1 & 1 & a \\
1 & 2 & b \\
1 & 3 & c
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & a \\
0 & 1 & b-a \\
0 & 2 & c-a
\end{bmatrix}
&& R_2\gets R_2-R_1, R_3\gets R_3-R_1\\
&\to
\begin{bmatrix}
1 & 1 & a \\
0 & 1 & b-a \\
0 & 0 & a-2b+c
\end{bmatrix}
&& R_3\gets R_3-2R_2
\end{align}
Can you find $a,b,c$ so that the system has no solution?

Note that the zero vector is in the span of any set of vectors, so it's definitely not a candidate. One can try an educated guess, which however would be out of the question in case we are in $\mathbb{R}^{1000}$ instead of $\mathbb{R}^3$.
Another fact that you can consider is that at least one of the vectors in the standard basis is not in the span, because otherwise this set would be a generating set and, its cardinality being less than the dimension, we'd get a contradiction.
This says that in the above elimination argument we are sure that one among $a,b,c$ can be chosen to be $1$ and the others to be $0$. But we don't know which one at the beginning. For instance, if you are given $(2,1,1)$ as the second vector, instead of $(1,2,3)$, the choice $a=1$, $b=c=0$ would not be good.
