Chapter 4 - Vector Spaces - 4.4 Spanning Sets and Linear Independence - 4.4 Exercises - Page 178: 38

$S$ is not linearly independent.

It is eas to note that $$(1,5,-3)=(1,0,0)+\frac{5}{4}(0,4,0)+\frac{1}{2}(0,0,-6).$$ Then we have the nontrivial combination $$(1,5,-3)-(1,0,0)-\frac{5}{4}(0,4,0)-\frac{1}{2}(0,0,-6)=(0,0,0,0).$$ Which means that $S$ is not linearly independent.