Answer
See answer below
Work Step by Step
We are given $S=\{(6,-3,2);(1,1,1),(1,-8,-1)\}$ in $R^3$
Obtain the matrix $\det A=\begin{bmatrix}
6 & 1 & 1\\
-3 & 1 & -8\\
2 & 1 & -1
\end{bmatrix}=6\begin{vmatrix}
1 & -8\\
1 & -1
\end{vmatrix}-\begin{vmatrix}
-3 & -8\\
2 & -1
\end{vmatrix}+\begin{vmatrix}
-3 & 1\\
2 & 1
\end{vmatrix}=6[-1-(-8)]-[3-(-16)]+(-3-2)=18 \ne 0$
Since $\det (A) \ne 0$, $S$ is linearly independent in $R^3$.
Hence, $S$ is a basic for $R^3$ and therefore $S$ spans $R^3$
