The direction cosines of the line $\frac{x-1}{l}=\frac{y+1}{l+1}=\frac{z-1}{l}$

If $⟨l,m,n⟩$ are direction cosines of a line then $l^2+m^2+n^2=1$

Hence,  $l^2+{(l+1)}^2+l^2=1$

It implies that  $3l^2+2l=0\Rightarrow l(3l+2)=0\Rightarrow l=0,-\frac{2}{3}$

Therefore, the direction cosines are $\boxed{⟨-\frac{2}{3},\frac{1}{3},-\frac{2}{3}⟩}$