Consider the matrix equation
$ k_1X_1+ k_2X_2+ k_3X_3+ k_4X_4=0 \\ \; \\ \; \\ \therefore k_1[1,2,1,0] k_2[1,3,1,2]+ k_3[4,2,1,0] + k_4[6,1,0,1] =0 \\ \; \\ \; \\ \therefore k_1+k_2+4k_3+6k_4= 0 \\ 2k_1+3k_2+2k_3+k_4= 0 \\ k_1+k_2+k_3+0k_4= 0 \\ 0k_1+k_2+0k_3+k_4= 0 $

Above equations can be written in matrix form as follows -

$ \left[ \begin{array}{cccc} 1 & 1 & 4 & 6 \\ 2 & 3 & 2 & 1 \\ -1 & 1 & 1 & 0 \\ 0 & 2 & 0 & 1 \end{array}\right] \left[ \begin{array}{cccc} k_1 \\ k_2 \\ k_3 \\ k_4 \end{array}\right] \left[ \begin{array}{cccc} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ R_2 \rightarrow R_2-2R_1 \; , \; R_3 \rightarrow R_3+R_1 \\ \; \\ \; \\ \left[ \begin{array}{cccc} 1 & 1 & 4 & 6 \\ 0 & 1 & -6 & -11 \\ 0 & 2 & 5 & 6 \\ 0 & 2 & 0 & 1 \end{array}\right] \left[ \begin{array}{cccc} k_1 \\ k_2 \\ k_3 \\ k_4 \end{array}\right] \left[ \begin{array}{cccc} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ R_3 \rightarrow R_3-R_4 \\ \; \\ \; \\ \left[ \begin{array}{cccc} 1 & 1 & 4 & 6 \\ 0 & 1 & -6 & -11 \\ 0 & 0 & 5 & 5 \\ 0 & 2 & 0 & 1 \end{array}\right] \left[ \begin{array}{cccc} k_1 \\ k_2 \\ k_3 \\ k_4 \end{array}\right] \left[ \begin{array}{cccc} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ R_3 \rightarrow R_3(1/5) \\ \; \\ \; \\ \left[ \begin{array}{cccc} 1 & 1 & 4 & 6 \\ 0 & 1 & -6 & -11 \\ 0 & 0 & 1 & 1 \\ 0 & 2 & 0 & 1 \end{array}\right] \left[ \begin{array}{cccc} k_1 \\ k_2 \\ k_3 \\ k_4 \end{array}\right] \left[ \begin{array}{cccc} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ R_2 \rightarrow R_2 + 6R_3 \\ \; \\ \; \\ \left[ \begin{array}{cccc} 1 & 1 & 4 & 6 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 1 \\ 0 & 2 & 0 & 1 \end{array}\right] \left[ \begin{array}{cccc} k_1 \\ k_2 \\ k_3 \\ k_4 \end{array}\right] \left[ \begin{array}{cccc} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ R_4 \rightarrow R_4 - 2R_2 \\ \; \\ \; \\ \left[ \begin{array}{cccc} 1 & 1 & 4 & 6 \\ 0 & 1 & 0 & -5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 11 \end{array}\right] \left[ \begin{array}{cccc} k_1 \\ k_2 \\ k_3 \\ k_4 \end{array}\right] \left[ \begin{array}{cccc} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] $

Note: This question seems wrong as $k_4$ is becoming equal to zero because of which $k_3$ = $k_2$ = $k_1$ = 0 . Hence vectors are linearly independent. But they’ve asked to prove that they are linearly dependent.