Added. Okay, let's clarify this, since you express confusion. If we multiply out $Ax$ and set it equal to $v$, we get: $\left(\begin{array}{c} x_1 + x_3 \\ -x_2 + 2x_3\\ x_1 - x_2 + 3x_3 \end{array}\right) = \left(\begin{array}{c}a\\b\\c\end{array}\right),$ which leads to the system of linear equations in three unknowns: $\begin{array}{rcccccl} x_1 & & & + & x_3 & = & a\\ & - & x_2 & + & 2x_3 & = & b\\ x_1 & - & x_2 & + & 3x_3 & = & c \end{array}$
The question is: for what vectors $v$ does this system have a solution $x$?
For example, if $v =\left(\begin{array}{c}0\\0\\0\end{array}\right)$ then the system has a solution (namely, $x_1=x_2=x_3=0$). If $v = \left(\begin{array}{c}1\\2\\3\end{array}\right)$ then the system has a solution (for example, $x_1 = 1$, $x_2=-2$, $x_3=0$; or $x_1=0$, $x_2=0$, $x_3=3$; or any of a number of others, but at least one). On the other hand, if $v=\left(\begin{array}{c}0\\1\\0\end{array}\right)$ then the system has no solutions: from $x_1+x_3=0$ we get $x_1=-x_3$. Substituting into the third equation, we would get $-x_2 + 2x_3 = 0$, so $x_2 = 2x_3$; but if we substitute that into the second equation, we get $-2x_3+2x_3 = 1$, which is impossible. So for this $v$, there are no solutions.
So... for some choices of $v$, the system has a solution. For other choices of $v$, the system has no solutions.
The question is: what are the choices of $v$ that will give you a system with a solution?
You are being asked to find the values of $a$, $b$, and $c$ for which the system $\begin{array}{rcccccl} x_1 & & & + & x_3 & = & a\\ & - & x_2 & + & 2x_3 & = & b\\ x_1 & - & x_2 & + & 3x_3 & = & c \end{array}$ has a solution. I hope you are familiar with solving systems of equations by Gaussian elimination, because that is what I'm going to do next. The augmented matrix of this system is just: $\left(\begin{array}{crr|c} 1 & 0 & 1 & a\\ 0 & -1 & 2 & b\\ 1 & -1 & 3 & c \end{array}\right)$ (that is, it is nothing more than $(A|v)$). We need to know when this system is consistent. If $a$, $b$, and $c$ were numbers, then Gaussian elimination is a recipe that tells us how to figure that out.
So then let's pretend they are specific numbers, and see where that leads us! We subtract the first row from the third row, and we obtain $\left(\begin{array}{rrr|c} 1 & 0 & 1 & a\\ 0 & -1 & 2 & b\\ 0 & -1 & 2 & c-a \end{array}\right).$ And now, subtracting the second row from the third row, we get $\left(\begin{array}{rrr|c} 1 & 0 & 1 & a\\ 0 & -1 & 2 & b\\ 0 & 0 & 0 & c-a-b \end{array}\right).$
Now... under what conditions would you know this system has no solutions? Under what conditions will it certainly have solutions?