Solution
= Solution
With $a=1$, $b=2$, $c=2$, substitute into the matrix equations. The equations reduce to $x+2z=0$ and $3y+z=0$, with the latter repeated. Taking $z=3t$ gives the complete <null space>
$$
\boxed{(x,y,z)=t(-6,-1,3),\qquad t\in\mathbb R.}
$$
The two independent rows impose two constraints on three unknowns, leaving exactly one free parameter. This also includes the zero solution at $t=0$.