Linear independence

(a) The system $\lambda_1(1,0,0)+\lambda_2(0,1,0)+\lambda_3(0,0,1)=0$ gives $\lambda_1=\lambda_2=\lambda_3=0$. Independent.

(b) The second vector is twice the first, so $2(1,2,3)-(2,4,6)=0$ is a nontrivial relation. Dependent.

(c) The third vector is the sum of the first two: $(1,1,1)=(1,0,1)+(0,1,0)$. Dependent.

(d) Solve $\lambda_1(1,1,0)+\lambda_2(0,1,1)+\lambda_3(1,0,1)=0$:
$$\lambda_1+\lambda_3=0,\quad \lambda_1+\lambda_2=0,\quad \lambda_2+\lambda_3=0.$$
From the first, $\lambda_3=-\lambda_1$; from the second, $\lambda_2=-\lambda_1$; the third gives $-\lambda_1-\lambda_1=0$, so $\lambda_1=0$ and then $\lambda_2=\lambda_3=0$. Independent.

Conclusion. (a), (d) are independent; (b), (c) are dependent.