Determining that this basis is linearly independent with a variable 
Have the basis
$$B = \{ (1,2,0) , (1,1,1) , (1,a,0) , (0,0,a) \}$$
Explain why doesn't this basis have a dimension of $4$.

The only way would be, I guess, that it is linearly dependent, so it should have less than $4$ vectors, thus having a lower dimension than $4$.
Since $a$ is free, let's say it is $a = 0$. We'd have
$$B = \{ (1,2,0) , (1,1,1) , (1,0,0) , (0,0,0) \}$$
Clearly $(0,0,0)$ is redundant, so it is dependent.
Is that reasoning right? I'm just not sure if it's enough by simply setting $a = 0$.

Hint: to determine whether vectors $\vec v_1, \vec v_2, \vec v_3, \vec v_4$ are linearly (in)dependent, solve $x_1\vec v_1+x_2 \vec v_2+x_3 \vec v_3+x_4 \vec v_4= \vec 0$.
If the only solution to this system of equations is $x_1=x_2=x_3=x_4=0$ (trivial solution), then the four vectors are linearly independent; otherwise, they're linearly dependent.
Solving $x_1\vec v_1+x_2 \vec v_2+x_3 \vec v_3+x_4 \vec v_4= \vec 0$, we get the following system:
$$\begin{bmatrix} 1 &1 &1 &0 \\ 2 & 1 & a & 0 \\ 0&1&0&a\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\ x_3 \\ x_4\end{bmatrix}=\begin{bmatrix} 0\\ 0 \\ 0 \\ 0\end{bmatrix}.$$
Solve this (Gaussian elimination) and (hopefully) deduce that there exists a non-trivial solution to this, meaning the four vectors are linearly dependent, so they can't possibly form a basis  of $\mathbb{R^3}$(or a cheat is to say that we have fewer equations than unknowns, so there must be a non-trivial solution to the system).
