$Let \ x_1=(1,2,-1,0), \ x_2=(1,3,1,3), \ x_3=(4,2,1,-1), \ x_4=(6,1,0,-5) \ and \ k_1, k_2, k_3, k_4 \ \in \ R \ such \ that \ k_1x_1+k_2x_2+k_3x_3+k_4x_4 =0$ $so, \ k_1\begin{bmatrix} 1\2 \-1 \ 0 \end{bmatrix}+ k_2 \begin{bmatrix} 1\3 \1 \3 \end{bmatrix}+ k_3 \begin{bmatrix} 4\2 \1 \-1 \end{bmatrix}+ k_4 \begin{bmatrix} 6\1 \0 \-5 \end{bmatrix}=0 $ $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+3k_2-k_3-5k_4=0 $ This is a homogeneous linear system of equation. The co-efficient matrix A is $A=\begin{bmatrix} 1 &1 &4 &6 \2 &3 &2 &1 \-1 &1 &1 &0 \0 &3 &-1 &-5 \end{bmatrix}$ $R_2-2R_1, \ \ R_3+R_1 $ $A\sim \begin{bmatrix} 1 &1 &4 &6 \0 &1 &-6 &-11 \0 &2 &5 &6 \0 &3 &-1 &-5 \end{bmatrix} $ $R_4-R_2$ $A\sim \begin{bmatrix} 1 &1 &4 &6 \0 &1 &-6 &-11 \0 &2 &5 &6 \0 &2 &6 &6 \end{bmatrix}$ $$R_4-R_3$ $A\sim \begin{bmatrix} 1 &1 &4 &6 \0 &1 &-6 &-11 \0 &2 &5 &6 \0 &0 &0 &0 \end{bmatrix}$ 3 non-zero rows. ∴ Rank of matrix A=3, But, number of variables, n=4 → Rank of A ∴ system has infinite non-trivial solutions for $k_1, k_2, k_3, k_4.$

Hence, (1,2-1,0), (1,3,1,3),(4,2.1,-1) & (6,1,0,-5) are linealy dependent vectors.