Consider the matrix equation $k_1 x_1+k_2 x_2+k_3 x_3=0 \\ \; \\ \; \\ \therefore k_1 [1,3,4,2]+k_2 [3,-5,2,6]+k_3 [2,-1,3,4]=0 \\ \; \\ \; \\ \therefore k_1+3k_2+2k_3=0 \; , 3k_1-5k_2-k_3=0, \; 4k_1+2k_2+3k_3=0, \; 2k_1+6k_2+4k_3=0 $

which can be written in matrix form as-

$ \left[ \begin{array}{ccc} 1 & 3 & 2 \\ 3 & -5 & -1 \\ 4 & 2 & 3 \\ 2 & 6 & 4 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2-3R_1 \; , \; R_3 \rightarrow R_3 - 4R_1 \; , \; R_4 \rightarrow R_4 - 2R_1 \\ \; \\ \left[ \begin{array}{ccc} 1 & 3 & 2 \\ 0 & -14 & -7 \\ 0 & -10 & -5 \\ 0 & 0 & 0 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_2 \rightarrow R_2 \times (-1/7) \; , \; R_3 \rightarrow R_3 \times (-1/5) \\ \; \\ \left[ \begin{array}{ccc} 1 & 3 & 2 \\ 0 & 2 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ R_3 \rightarrow R_3 - R_2 \\ \; \\ \left[ \begin{array}{ccc} 1 & 3 & 2 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \left[ \begin{array}{ccc} k_1 \\ k_2 \\ k_3 \end{array}\right] \;=\; \left[ \begin{array}{ccc} 0 \\ 0 \\ 0 \\ 0 \end{array}\right] \\ \; \\ \; \\ \; \\ \therefore k_1+3k_2+2k_3=0,2k_2+k_3=0 \\ \; \\ \; \\ If we put k_3=t,k_2=-k_3/2=-t/2 \\ \; \\ \; \\ k_1+3k_2+2k_3=0 \\ \; \\ \; \\ \therefore k_1-\dfrac{3t}{2}+2t=0 \; \; \therefore k_1=-\dfrac{t}{2} \\ \; \\ \; \\ \therefore -\dfrac{t}{2} x_1 -\dfrac{t}{2} x_2 + tx_3 = 0 $

Since, $k_1,k_2,k_3$ are not all zero,the vectors are linearly independent and $x_1=-x_2+2x_3.$