Answer
a. no, 3
b. infinitely many
c. yes, $v_{1}+v_{2}=w$
Work Step by Step
a.
The set $S=\{v_{1},v_{2},v_{3}\}$ contains the three given vectors.
$w$ is not equal to any of them, so $w\not\in S$.
b. Span$ S=\{av_{1}+bv_{2}+cv_{3}, a,b,c\in \mathbb{R}\}$
has infinitely many vectors
(there are infinitely many ways of choosing a,b, and c)
c.
If $ w\in$Span$ S$, then there exist $a,b,c\in \mathbb{R}$
such that $w=av_{1}+bv_{2}+cv_{3}.$
$a, b,$ and $c$ are solutions (if they exist)
to a linear system with the augmented matrix
$\left[\begin{array}{lllll}
1 & 2 & 4 & | & 3\\
0 & 1 & 2 & | & 1\\
-1 & 3 & 6 & | & 2
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\leftarrow R_{3}+R_{1}.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & 4 & | & 3\\
0 & 1 & 2 & | & 1\\
0 & 5 & 10 & | & 5
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\leftarrow (\frac{1}{5}R_{3}).
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & 4 & | & 3\\
0 & 1 & 2 & | & 1\\
0 & 1 & 2 & | & 1
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\leftarrow R_{3}-R_{2}.
\end{array}\right.$
$\left[\begin{array}{lllll}
1 & 2 & 4 & | & 3\\
0 & 1 & 2 & | & 1\\
0 & 0 & 0 & | & 0
\end{array}\right]$
(consistent, solutions exist)
We take any $c\in \mathbb{R},$
back substituing, $b=1-2c$
back substituing, $a=3-2(1-2c)-4c=1,$
So, there exist a,b,c such that $ w\in$Span$ S$.
(For example, c=0$\Rightarrow v_{1}+v_{2}=w$)