I don't understand what you've been doing. Let's call $u$ and $v$ the vectors from $S$ and $w$ the one from (a). What do you want to know is if there exists real numbers $x$ and $y$ such that
$$
w = xu + y v \ .
$$
Right?
But this is a system of linear equations, with four equations (right) and just two unknowns ($x$ and $y$). This one:
$$
\begin{pmatrix}
-42 \\ 113 \\ -112 \\ -60
\end{pmatrix}
=
x
\begin{pmatrix}
6 \\ -7 \\ 8 \\ 6
\end{pmatrix}
+
y
\begin{pmatrix}
4 \\ 6 \\ -4 \\ 1
\end{pmatrix}
$$
Which you can write as
$$
\begin{pmatrix}
6 & 4 & \vert & -42 \\
-7 & 6 &\vert & 113 \\
8 & -4 &\vert & -112 \\
6 & 1 & \vert & -60
\end{pmatrix}
$$
If I'm not wrong, this time this system has no solution at all (and so, $w$ is NOT a linear combination of $u$ and $v$), but you don't have to think that, just because a system of linear equations has more unknowns than equations this necessarily means that it has no solutions. For instance, this one
$$
\begin{align}
x + y &=& 1 \\
x + y &=& 1 \\
x + y &=& 1
\end{align}
$$
Has two unknowns and three equations, but an infinite number of solutions. "Oh, but it's always the same equation, so it doesn't count". -Is that what you're thinking? Well, try to solve this one:
$$
\begin{align}
x + y &=& 1 \\
x - y &=& 0 \\
2x &=& 1
\end{align}
$$