Answer
See answer below
Work Step by Step
We are given:
$\{(1,2,1),(3,-1,2),(1,1,-1)\}$
Then we have:
$ \begin{vmatrix}
1 & 3 & 1\\
2 & -1 & 1 \\
1 & 2 & -1
\end{vmatrix}=\begin{vmatrix}
1 & 1\\
2 & -1
\end{vmatrix}-3\begin{vmatrix}
2 & 1 \\
1 & -1
\end{vmatrix}+\begin{vmatrix}
2 & -1\\
1 & 2
\end{vmatrix}=(1-2)-3(-2-1)+(4+1)=13 \ne0 $
We can see that the set is linearly independent in $R^3$
Since $dim R^3=3$, the set $\{(1,2,1),(3,-1,2),(1,1,-1)\}$ is a basic for $R^3$ (Theorem 4.6.10)
