Here’s what this particular question is asking:
Suppose that $A$ is a $3\times 3$ matrix such that the equation $Ax=\left[\matrix{1\\1\\1}\right]\tag{1}$ has a solution; is it then necessarily true that the equation $Ax=\left[\matrix{1\\2\\1}\right]\tag{2}$ also has a solution?
The answer is no, because when $A=\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]\;,$ equation $(1)$ has a solution, but equation $(2)$ does not.
$\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]\left[\matrix{1\\0\\0}\right]=\left[\matrix{1\\1\\1}\right]\;,$ but
$\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]\left[\matrix{x\\y\\z}\right]=\left[\matrix{x+y+z\\x+y+z\\x+y+z}\right]\;,$ and no matter what $x,y$, and $z$ you choose, $\left[\matrix{x+y+z\\x+y+z\\x+y+z}\right]\ne\left[\matrix{1\\2\\1}\right]\;,$ because all three of its components are the same.
In general, if your asked to prove or disprove a statement of the form $\text{if }P\text{ is true then }Q\text{ is true}\;,$ you first have to decide whether it is true: if the mathematical objects mentioned in the statement are chosen in such a way that $P$ is true, does that automatically make $Q$ true as well? If the answer is yes, you’ll need to find a reason, a logical argument leading from the truth of $P$ to the truth of $Q$. For example, suppose that the statement is:
if the $3\times 3$ matrix $A$ is invertible, then the equation $Ax=\left[\matrix{a\\b\\c}\right]$ has a solution for every possible choice of real numbers $a,b,c$,
you can show that it’s true by pointing to the solution $x=Ix=(A^{-1}A)x=A^{-1}(Ax)=A^{-1}\left[\matrix{a\\b\\c}\right]\;.$
If the statement is not always true, your job is in some ways easier: you just have to demonstrate one example in which $P$ is true and $Q$ is not. That’s what I did in the specific example that you gave: the matrix $A=\left[\matrix{1&1&1\\1&1&1\\1&1&1}\right]$ shows that equation $(1)$ can have a solution without $(2)$ also having a solution.