Step 1: Interpreting the given solution form:
It is given that $X = [l, m, 0]^T$ is a solution of $AX = O$ for all non-zero $l, m$.
This implies that every vector lying in the xy-plane (i.e., vectors with $z = 0$) satisfies the equation.
Hence, the entire xy-plane is contained in the null space (kernel) of matrix $A$.
Step 2: Finding the nullity of $A$:
The xy-plane is a two-dimensional subspace, and vectors such as
\[
\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}
\]
form its basis.
Therefore, the null space has dimension $2$, i.e.,
\[
\text{nullity}(A) = 2.
\]
Since $A$ is a $3 \times 3$ matrix, the nullity cannot exceed $3$. If it were $3$, all vectors would be solutions, which contradicts the given condition.
Step 3: Applying the Rank-Nullity Theorem:
For a matrix with $n$ columns,
\[
\text{rank}(A) + \text{nullity}(A) = n.
\]
Here, $n = 3$, so
\[
\text{rank}(A) + 2 = 3 \Rightarrow \text{rank}(A) = 1.
\]
Step 4: Verification using column interpretation:
Since every vector of the form $[l, m, 0]^T$ is a solution, we have
\[
A \begin{pmatrix} l \\ m \\ 0 \end{pmatrix} = l \cdot C_1 + m \cdot C_2 = 0,
\]
where $C_1$ and $C_2$ are the first two columns of $A$.
This is possible only if both $C_1$ and $C_2$ are zero vectors.
Thus, $A$ must be of the form
\[
\begin{pmatrix}
0 & 0 & c_1 \\
0 & 0 & c_2 \\
0 & 0 & c_3
\end{pmatrix},
\]
which clearly has rank $1$ (provided it is not the zero matrix).
Final Conclusion:
Therefore, the rank of matrix $A$ is
\[
\boxed{1}.
\]