## Elementary Linear Algebra 7th Edition

$S$ is not a basis for $R^3$.
The set $S=\{(0,0,0),(1,5,6),(6,2,1)\}$ for $R^{3}$ is not a basis for $R^3$ because $S$ is not linearly independent set of vectors. Indeed, assume that $$a(0,0,0)+b(1,5,6)+c(6,2,1)=(0,0,0), \quad a,b,c\in R.$$ 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.