Consider the matrix equation $k_1X_1+k_2X_2+k_3X_3=0 \cdots (A)$

$k_1[1,3,4,2]+k_2[3,-5,2,6]+k_3[2,-1,3,4]=[0,0,0,0]$

$k_1+3k_2+2k_3=0 \cdots (1)$

$3k_1-5k_2-1k_3=0 \cdots (2);\\ 4k_1+2k_2+3k_3=0 \cdots (3);\\ 2k_1+6k_2+4k_3=0 \cdots (4)$

Write equation 1,2,3,4  in matrix form,

$\begin{bmatrix} 1 & 3 & 2 \\[0.3em] 3 & -5 & -1 \\[0.3em] 4 & 2 & 3 \\[0.3em] 2 & 6&4 \end{bmatrix}. \begin{bmatrix} k_1 \\[0.3em] k_2 \\[0.3em] k_3 \end{bmatrix}= \begin{bmatrix} 0 \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] 0 \end{bmatrix}$

By $R_2-3R_1,\ R_3-4R_1,\ R_4-2R_1$

$\begin{bmatrix} 1 & 3 & 2 \\[0.3em] 0 & -14 & -7 \\[0.3em] 0 & -10 & -5 \\[0.3em] 0 & 0 &0 \end{bmatrix}. \begin{bmatrix} k_1 \\[0.3em] k_2 \\[0.3em] k_3 \end{bmatrix}= \begin{bmatrix} 0 \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] 0 \end{bmatrix}$

By $(-1/7)R_2,\ (-1/5)R_3$

$\begin{bmatrix} 1 & 3 & 2 \\[0.3em] 0 & 2 & 1 \\[0.3em] 0 & 2 & 1 \\[0.3em] 0 & 0&0 \end{bmatrix}. \begin{bmatrix} k_1 \\[0.3em] k_2 \\[0.3em] k_3 \end{bmatrix}= \begin{bmatrix} 0 \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] 0 \end{bmatrix}$

By $R_3-R_2$

$\begin{bmatrix} 1 & 3 & 2 \\[0.3em] 0 & 2 & 1 \\[0.3em] 0 & 0 &0 \\[0.3em] 0 &0 &0 \end{bmatrix}. \begin{bmatrix} k_1 \\[0.3em] k_2 \\[0.3em] k_3 \end{bmatrix}= \begin{bmatrix} 0 \\[0.3em] 0 \\[0.3em] 0 \\[0.3em] 0 \end{bmatrix}$

$k_1+3k_2+2k_3=0 \cdots (5)\\ 2k_2+k_3=0 \cdots (6)$

$Put\ 2t=k_3\ in (6), we\ get, 2k_2+2t=0; k_2=-t$

Substitute, $ k_3 \ and\ k_2 \ in\ (5), k_1-3t+4t=0; k_1=-t$

Substitute, $k_1, k_2, k_3 \ in \ (4);\ -tX_1-tX_2+2tX_3=0$

$Since \ k_1, k_2, k_3\ are\ not\ zero,\ hence, the\ vectors\ are\ linearly \ dependent\ and\ their\ relationship\ is\ given\ by\ X_1=-X_2+2X_3$