Given a matrix A = \(\begin{bmatrix} 2 & 4 & 1 \ 1 & 2 & 1 \ 3 & 6 & 2 \end{bmatrix}\), find the null space.

Step 1 ：Step 1: Find the row reduced echelon form (RREF) of the matrix. \[ RREF = \begin{bmatrix} 1 & 2 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0 \end{bmatrix} \]

Step 2 ：Step 2: From RREF, the system of equations is \[\begin{cases} x + 2y = 0 \ z = 0 \end{cases}\]

Step 3 ：Step 3: Solving the system, we get \[\begin{cases} x = -2y \ z = 0 \end{cases}\]