Determining for which parameter values a set: spans a vector space, is linearly independent, a base for a vector space. Let $V = \mathbb{R}^{3}$
and $v_1 = \begin{bmatrix}1\\ a\\ 1\end{bmatrix}$, $v_2 = \begin{bmatrix}1\\ a+1\\ 1\end{bmatrix}$, $v_3 = \begin{bmatrix}3\\ a+2\\ 3\end{bmatrix}$ $a\in \mathbb{R}$
Let $S = \left \{ v_1, v_2, v_3 \right \}$
For which $a\in \mathbb{R}$ :
$S$ spans $V$
$S$ is linearly independent
$S$ is a base for $V$.
So here's how I look at it: I can immidiately tell that $v_1, v_2$ are linearly independent, because each one of their coordinates is the same except for the middle one, and the equation $a = a+1$ has no solution -  therefore for each $a\in \mathbb{R}$ $v_1$ will no be equal to $v_2$.
I run the same procces between $v_1$ and $v_3$ and between $v_2$ and $v_3$ and I find out that they're linearly independent when $a \neq 1, \frac{1}{2}$
Since I found that these 3 vectors are linearly independent for $a \neq 1, \frac{1}{2}$ I know that $S$ is a base for $V$,  because $\left | S \right | = dimV$
And therfore $S$ ofcourse spans $V$ for these values.
However, this is an easy case of "look and understand" (Assuming that I got it right)..
I would like to to get a more direct and general approach to attack this kind of question. thank you!

As explained in the comments, it is not sufficient to check for pairwise linear independence to conclude linear independence of the set ($S$).

and $v_1 = \begin{bmatrix}1\\ a\\ 1\end{bmatrix}$, $v_2 = \begin{bmatrix}1\\ a+1\\ 1\end{bmatrix}$, $v_3 = \begin{bmatrix}3\\ a+2\\ 3\end{bmatrix}$ $a\in \mathbb{R}$

Note that for any $a \in \mathbb{R}$, you have:
$$(1+2a)\begin{bmatrix}1\\ a\\ 1\end{bmatrix}
+(2-2a)\begin{bmatrix}1\\ a+1\\ 1\end{bmatrix}
=\begin{bmatrix}3\\ a+2\\ 3\end{bmatrix}$$
which means that $v_1$, $v_2$ and $v_3$ are always linearly dependent.

I would like to to get a more direct and general approach to attack this kind of question. thank you!

For three vectors in $\mathbb{R^3}$, a handy equivalent property uses the determinant: the matrix formed by the three vectors has a non-zero determinant if and only if the vectors are linearly independent.
You can easily verify that for all $a \in \mathbb{R}$ (the first and last rows are identical):
$$\begin{vmatrix}
1 & 1 & 3 \\
a & a+1 & a+2 \\
1 & 1 & 3
\end{vmatrix}=0$$

Hint:
Let be $M\in\mathbb R^{3\times3}$ the matrix with columns $v_1$, $v_2$ and $v_3$. $S$ is linearly independent if and only if $M$ has full rank. The row rank of $M$ is obvious.