Why not just reduce the matrix ? its not more work and often makes things easier.
\begin{bmatrix}
1&0&0&0
\\0&1&0&0
\\0&0&1&0
\end{bmatrix}
1) We see that the row and column rank (they are always equal) are $3$.
The nullity $=n-$rank$=4-3=1$.
To find the null space we have to solve $A\mathbf{x}=0$, and this is easy now in row reduced form $\mathbf{x}=\begin{bmatrix}
0\\0\\0\\a
\end{bmatrix}$.
2) The row space has dimension $3$ as mentioned, for the basis one can take:
$$(1,0,0,0)$$
$$(0,1,0,0)$$
$$(0,0,1,0)$$
Or one could take the rows of the original matrix, since the rank is $3$.
3) The column rank is also $3$ row reduction has not changed the column vectors, just expressed them in a different basis so a basis for the column space will be the first $3$ vectors of the original matrix (corresponding to the pivot position of the reduced matrix):
$$\begin{bmatrix}
1\\3\\-1
\end{bmatrix},
\begin{bmatrix}
1\\-1\\2
\end{bmatrix},
\begin{bmatrix}
-1\\2\\-4\\
\end{bmatrix}. $$
This matrix reduces, through row reduction, to $\begin{pmatrix}1 & -1 & 0 & 0 \\ 0 & -1 & -4 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$. The row space has the three basis vectors, (1, -1, 0, 0), (0, -1, -4, 0), and (0, 0, 0, 1), so dimension 3. The column space has the three basis vectors (1, 0, 0, 0), (-1, -1, 0, 0), and (0, 0, 1, 0). (0, -4, 0, 0) is not independent because (0, -4, 0, 0)= 4(-1, -1, 0, 0)+ 4(1, 0, 0, 0). The column space has dimension 3. That's always true- the dimension of the row space of a matrix is equal to the dimension of the column space".
(x, y, z, t) is in the "null space" if and only if $\begin{pmatrix}11 & -2 & 36 & 2 \\ -2 & 1 & -4 & 0 \\ 3 & 0 & 12 & 1 \\ 1 & -1 & 0 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z\\ t\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \\ 0\end{pmatrix}$. That is equivalent to the four equations 11x- 2y+ 36z+ 2t= 0, -2x+ y- 4z= 0, 3x+ 12z+ t= 0, x- y= 0. From x- y= 0, of course, y= x so the other three equations can be written 9x+ 36z+ 2t= 0, -x- 4z= 0, and 3x+ 12z+ t= 0. From -x- 4z= 0, x= -4z so the other two equations can be written 2t= 0, 0= 0, and t= 0. Clearly t= 0 but we cannot solve for numerical values of x, y, and z. We can say that (x, y, z, t)= (-4z, -4z, z)= z(-4, -4, 1) where z can be any number. That is a basis for the null space is {(-4, -4, 1)} and the dimension of the null space (the "nullity") is 1.
Note that the dimension of the null space, 1, plus the dimension of the row space, 1+ 3= 4, the dimension of the whole space. That is always true. After finding a basis for the row space, by row reduction, so that its dimension was 3, we could have immediately said that the column space had the same dimension, 3, and that the dimension of the null space was 4- 3= 1 without any more computation.
Best Answer
Since $Au=0$, $Aw=2w$ and $Az=3z$ by definition, you know that $2w$ and $3z$ belong to the column space of $A$, so $w$ and $z$ do as well. Since $\{w,z\}$ is a linearly independent set, you know that the column space has dimension at least $2$.
Moreover $u$ belongs to the null space, therefore the null space has dimension at least $1$. Since $3=1+2$, the rank-nullity theorem says that
A basis for the former is $\{u\}$; a basis for the latter is $\{w,z\}$.
How can we solve $Ax=w+z$? Since $A$ is diagonalizable, having three distinct eigenvalues, we have $x=\alpha u+\beta w+\gamma z$; then $$ Ax=\alpha Au+\beta Aw+\gamma Az=2\beta w+3\gamma z $$ so the equality $$ 2\beta w+3\gamma z=w+z $$ implies $\beta=1/2$ and $\gamma=1/3$. You can choose $\alpha$ without restrictions.
Can you finish the $Ax=u$ part?