Additive Identity of a Vector Space Looking at the various axioms for vector spaces, I'm getting hung up on this one:    

Additive Identity
  The set $V$ contains an additive identity element, denoted by $0$, such that for any vector $v$ in V,     $0 + v = v$   and   $v + 0 = v$.    

It seems simple enough but in an example given -    
$V = \Bigg\{
\begin{bmatrix}
1+x\\2-x\\3+2x
\end{bmatrix}     
\Bigg| x \in \mathbb{R}\Bigg\}$    
And the additive identity is -
$0=\begin{bmatrix}
1\\2\\3
\end{bmatrix}$    
I'm confused by this as certainly, $v + 0 \neq v$.     
I'd appreciate any help in understanding this, thanks!

Presumably the vector space you are working in is $\mathbb{R}^3$, the set of ordered $3$-tuples of real numbers.  An element in $\mathbb{R}^3$ is often written in the form $\overrightarrow{v}=\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}$.
With $\overrightarrow{v}=\begin{bmatrix} v_1\\v_2\\v_3\end{bmatrix}$ and $\overrightarrow{u}=\begin{bmatrix}u_1\\u_2\\u_3\end{bmatrix}$, we define addition in $\mathbb{R}^3$ in the following way:
$\overrightarrow{v}+\overrightarrow{u} = \begin{bmatrix} v_1\\v_2\\v_3\end{bmatrix} + \begin{bmatrix} u_1\\u_2\\u_3\end{bmatrix} = \begin{bmatrix} v_1+u_1\\v_2+u_2\\v_3+u_3\end{bmatrix}$
I.e. we separately add each component.
Given this definition of addition, the only possible zero vector would be $\overrightarrow{0}=\begin{bmatrix}0\\0\\0\end{bmatrix}$

In your example, you have $V$ defined as $V=\left\{\begin{bmatrix}1+x\\2-x\\3+2x\end{bmatrix}~:~x\in\mathbb{R}\right\}$
Given a more exotic addition, I'll use the symbol $\oplus$, this could still be made into a group and possibly a vector space where addition is defined in the following way:
$\overrightarrow{v}\oplus \overrightarrow{u} = \begin{bmatrix} v_1\\v_2\\v_3\end{bmatrix} \oplus \begin{bmatrix} u_1\\u_2\\u_3\end{bmatrix} = \begin{bmatrix} v_1+u_1-1\\v_2+u_2-2\\v_3+u_3-3\end{bmatrix}$
In this case, as your problem suggests, $\begin{bmatrix}1\\2\\3\end{bmatrix}$ acts as an additive identity, I will denote as $id_\oplus$.  To see this, note that $\overrightarrow{x} \oplus id_\oplus = \begin{bmatrix}1+x\\2-x\\3+2x\end{bmatrix}\oplus\begin{bmatrix}1\\2\\3\end{bmatrix}=\begin{bmatrix}1+x+1-1\\2-x+2-2\\3+2x+3-3\end{bmatrix}=\begin{bmatrix}1+x\\2-x\\3+2x\end{bmatrix}=\overrightarrow{x}$
As for closure under addition, suppose $\overrightarrow{x} = \begin{bmatrix} 1+x\\2-x\\3+2x\end{bmatrix}$ and $\overrightarrow{y} = \begin{bmatrix} 1+y\\2-y\\3+2y\end{bmatrix}$.  Then we have $\overrightarrow{x}\oplus \overrightarrow{y} = \begin{bmatrix} 1+x+1+y-1\\ 2-x+2-y-2\\3+2x+3+2y-3\end{bmatrix}=\begin{bmatrix}1+(x+y)\\2-(x+y)\\3+2(x+y)\end{bmatrix}$, so $\overrightarrow{x}\oplus\overrightarrow{y}\in V$.
For additive inverse, we want to show that any $\overrightarrow{x}\in V$ has some "$\ominus \overrightarrow{x}$" such that $\overrightarrow{x}\oplus (\ominus \overrightarrow{x}) = id_\oplus$ (where here $id_\oplus$ is the additive identity for $\oplus$ as described above).
Let $\overrightarrow{x}=\begin{bmatrix}1+x\\2-x\\3+2x\end{bmatrix}$ and $\ominus\overrightarrow{x}=\begin{bmatrix}1+(-x)\\2-(-x)\\3+2(-x)\end{bmatrix}$.  Then indeed, $\overrightarrow{x}\oplus (\ominus \overrightarrow{x}) = \begin{bmatrix}1+x+1-x-1\\2-x+2+x-2\\3+2x+3-2x-3\end{bmatrix}=id_\oplus$, so additive inverses exist.
Since it is closed under addition, has an additive identity, and has additive inverses for each element, this forms a group.
Given a suitable definition for scalar multiplication (where it essentially ignores the 1,2,3 at the beginning), then we can even make this a vector space.
However, with addition as normally defined, our zero vector should still be the traditional one where each entry is zero.  As a result, since there does not exist any value of $x$ such that $\begin{bmatrix}1+x\\2-x\\3+2x\end{bmatrix}=\overrightarrow{0}$, it will not be a vector space.
We see also that given usual addition $V$ will not be a vector space since it is not closed under addition (how it is usually defined) or scalar multiplication (how it is usually defined).
