Chapter 4 - Vector Spaces - 4.5 Basis and Dimension - 4.5 Exercises - Page 188: 49

$S$ is not a basis for $R^3$.

The set$S=\{(0,0,0),(1,3,4),(6,1,-2)\}$ for $R^{3}$ is not a basis for $R^3$ because $S$ is not linearly independent set of vectors. Indeed, assume that $$a(0,0,0)+b(1,3,4)+c(6,1,-2)=(0,0,0), \quad a,b,c\in R.$$ One can see that the above combination is true for the choice $a\neq 0, b=c=0$. This means that $S$ is Linearly independent.