= Solution
Reducing the augmented matrix gives
$$
\operatorname{rref}(A\mid b)=\begin{pmatrix}1&0&1&0\\0&1&2&1/2\\0&0&0&0\end{pmatrix}.
$$
There is no contradictory row. Hence the system is \b[consistent], with <affine solution space>
$$
\boxed{(x,y,z)=(0,1/2,0)+t(-1,-2,1),\qquad t\in\mathbb R.}
$$
The three <planes> intersect in this common line. Their normals span a two-dimensional <row space>, leaving a one-dimensional common direction, the <null space>. More explicitly, five times the first equation minus eleven times the second plus twice the third is the zero equation, including the right-hand side; the third plane therefore contains the intersection of the first two.
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