Solution
= Solution
All three <linear equations> are multiples of the first, leaving the single constraint $x+y+z=0$. Choose two <real numbers> $s,t$ freely:
$$
\boxed{(x,y,z)=(s,t,-s-t),\qquad s,t\in\mathbb R}.
$$
A <basis> for this two-dimensional <linear subspace> is $(1,0,-1)^T,(0,1,-1)^T$.