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 $$S=\{(2,0,1),(2,-1,1),(4,2,0)\}.$$
(a) Consider $u\in R^3$ such that
$$u=(x,y,z)=a(2,0,1)+b(2,-1,1)+c(4,2,0), \quad a,b,c\in R.$$
Which yields the following system of equations
\begin{align*}
2a+2b+4c&=x\\
-b+2c&=y\\
a+b&=z.
\end{align*}
The coefficient matrix
$$\left[ \begin {array}{ccc} 2&2&4\\ 0&-1&2\\ 1&1&2\end {array}
\right] $$
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
$$a(2,0,1)+b(2,-1,1)+c(4,2,0)=(0,0,0), \quad a,b,c\in R.$$
Which yields the following system of equations
\begin{align*}
2a+2b+4c&=0\\
-b+2c&=0\\
a+b&=0.
\end{align*}
The coefficient matrix
$$\left[ \begin {array}{ccc} 2&2&4\\ 0&-1&2\\ 1&1&2\end {array}
\right] $$
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 $R^3$.
