Answer
(a) $S$ spans $R^3$.
(b) $S$ is linearly independent set of vectors.
(c) $W$ is a basis for $R^3$.
Work Step by Step
Let $S$ be given by $$S=\{(1,-5,4),(11,6,-1),(2,3,5)\}.$$
(a) Consider $u\in R^3$ such that
$$u=(x,y,z)=a(1,-5,4)+b(11,6,-1)+c(2,3,5), \quad a,b,c\in R.$$
Which yields the following system of equations
\begin{align*}
a+11b+2c&=x\\
-5a+6b+3c&=y\\
4a-b+5c&=z.
\end{align*}
The coefficient matrix
$$\left[ \begin {array}{ccc} 1&11&1\\ -5&6&3\\ 4&-1&5\end {array}
\right]
$$
has non zero determinant and hence there exist a unique solution for the above system and hence one can calculate $a,b,c$. Therefore, $S$ spans $R^3$.
(b) Assume that
$$a(1,-5,4)+b(11,6,-1)+c(2,3,5)=(0,0,0), \quad a,b,c\in R.$$
Which yields the following system of equations
\begin{align*}
a+11b+2c&=0\\
-5a+6b+3c&=0\\
4a-b+5c&=0.
\end{align*}
The coefficient matrix
$$\left[ \begin {array}{ccc} 1&11&1\\5&6&3\\ 4&-1&5\end {array} \right]
$$
has non zero determinant and hence there exist a unique solution for the above system, that is, the trivial solution $a=0,b=0,c=0$. Hence $S$ is linearly independent set of vectors.
(c) Since $W$ is linearly independent set and spans $R^3$, then it is a basis for $R^3$.
