Answer
$S$ is not a basis for $R^3$.
Work Step by Step
The set $S=\left\{\left(\frac{2}{3}, \frac{5}{2}, 1\right),\left(1, \frac{3}{2}, 0\right),(2,12,6)\right\}$
is a linearly dependent set of vectors. Indeed,
assume that
$$a\left(\frac{2}{3}, \frac{5}{2}, 1\right)+b\left(1, \frac{3}{2}, 0\right)+c(2,12,6)=(0,0,0), \quad a,b,c\in R.$$
Then, we have the following system of equations
\begin{align*}
\frac{2}{3}a+b+2c&=0\\
\frac{5}{2}a+\frac{3}{2}b+12c&=0\\
a+6c&=0.
\end{align*}
The augmented matrix of the above system is given by
$$\left[\begin{array}{cccc}
\frac{2}{3}&1&2\\
\frac{5}{2}&\frac{3}{2}&12
\\
1&0&6
\\
\end{array}\right]$$
and its determinant is $0$. Therefore, the above system has infinite number of solutions, that is, there is lots of non trivial solution. Then $S$ is linearly dependent set of vectors and then $S$ is not a basis for $R^3$.
