Answer
See below
Work Step by Step
Given: $\{(1,1,k),(0,2,k),(1,k,6)\} \in R^3$
The set is linearly independent if $\begin{vmatrix}
1 & 0 & 1\\
1 & 2 & k\\
k & k & 6
\end{vmatrix}=0\\
\rightarrow \det \begin{vmatrix}
1 & 0 & 1\\
1 & 2 & k\\
k & k & 6
\end{vmatrix}=\begin{vmatrix}
2 & k\\ k & 6
\end{vmatrix}-0\begin{vmatrix}1 & k\\
k & 6
\end{vmatrix}+\begin{vmatrix}
1 & 2 \\
k & k
\end{vmatrix}=(12-k^2)-0+(k-2k)=-(k^2-k+12)=-(k-3)(k+4)=0$
then $k=3$ or $k=-4$
Hence, the value of constant $k$ for which the vectors $\{(1,1,k),(0,2,k),(1,k,6)\}$ is linearly independent are $k=-4, k=3$
