Consider the matrix equation $k_1 x_1+k_2 x_2+k_3 x_3=0$

$ \therefore k_1 [3,1,1]+k_2 [2,0,-1]+k_3 [4,2,1]=0 \\ \; \\ \; \\ \therefore 3k_1+2k_2+4k_3=0, \; \; k_1+2k_3=0, \; \; k_1-k_2+k_3=0 $

which can be writtten in matrix form as-

$ \left[ \begin{array}{ccc} 3 & 2 & 4 \\ 1 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ R_1 \rightarrow R_1-2R_2 \; , \; R_3 \rightarrow R_3-R_2 \\ \; \\ \left[ \begin{array}{ccc} 1 & 2 & 0 \\ 1 & 0 & 2 \\ 0 & -1 & -1 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ R_2 \rightarrow R_2-R_1 \\ \; \\ \left[ \begin{array}{ccc} 1 & 2 & 0 \\ 0 & -2 & 2 \\ 0 & -1 & -1 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ R_2 \rightarrow R_2-2R_3 \\ \; \\ \left[ \begin{array}{ccc} 1 & 2 & 0 \\ 0 & 0 & 4 \\ 0 & -1 & -1 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ R_2 \rightarrow R_2 \times (1/4) \; , \; R_2 \longleftrightarrow R_3 \\ \; \\ \left[ \begin{array}{ccc} 1 & 2 & 0 \\ 0 & -1 & -1 \\ 0 & 0 & 1 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ R_2 \rightarrow R_2 + R_3 \\ \; \\ \left[ \begin{array}{ccc} 1 & 2 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ R_1 \rightarrow R_1 + 2R_2 \\ \; \\ \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ \therefore k_1=k_2=k_3=0 $

Hence, given vectors are linearly independent.