Answer
$S$ is a basis for $R^3$.
$$u=(8,3,8)=5(1,0,0)-5(1,1,0)+8(1,1,1).$$
Work Step by Step
The set $S=\{(1,0,0),(1,1,0),(1,1,1)\}$ is a linearly independent set of vectors. Indeed,
assume that
$$a(1,0,0)+b(1,1,0)+c(1,1,1)=(0,0,0), \quad a,b,c\in R.$$
Then, we have the following system of equations
\begin{align*}
a+b+c&=0\\
b+c&=0\\
c&=0.
\end{align*}
Solving the above equations, we find that $a=0,b=0,c=0$ and hence $S$ is linearly independent set of vectors. Now, since $R^3$ is vector space of dimension $3$, then by Theorem 4.12, $S$ is a basis for $R^3$.
To write $u=(8,3,8)$ as a linear combination of the vectors in $S$, suppose that
$$a(1,0,0)+b(1,1,0)+c(1,1,1)=(8,3,8)$$
Then, we have the following system of equations
\begin{align*}
a+b+c&=8\\
b+c&=3\\
c&=8.
\end{align*}
Solving the above equations, we find that $a=5,b=-5,c=8$. Hence
$$u=(8,3,8)=5(1,0,0)-5(1,1,0)+8(1,1,1).$$
