System of five linear equations but only one is knowing

System of five linear equations but only one is knowing

Let suppose that we have system of five linear equations with three unknow
$x,y,z$. We know that one of equation of this system is $x+y+z=3$ and
$(x,y,z)=(3,0,0),(x,y,z)=(0,3,0)$ are solutions for this system. Is it
true that:
$(x,y,z)=(0,2,1)$ isn't solution of the system
$(x,y,z)=(0,1,2)$ is solution of the system
$(x,y,z)=(3,3,0)$ isn't solution of the system
$(x,y,z) = (1,2,0)$ is solution of the system
As we can see in 3. we have $3+3+0 = 6 \neq 3$ so $(3,3,0)$ isn't solution
of this system. In 4. I can use simple lemma: if $x,y$ are solutions of
$AX=b$ then for any $\alpha \in \mathbb{R}$ $\alpha x + (1- \alpha)y$ is
also solution. For $\alpha = \frac{1}{3}$ we have
$(1,2,0)=\frac{1}{3}(3,0,0)+\frac{2}{3}(0,3,0)$ so it is solution of this
system.
But I have no idea for 1. and 2. I will grateful for your help.