You must have made a mistake somewhere. If you notice that $v_2=v_1+v_3$, then you see immediately that the system will not have unique solution.
If I start with the same system
$$\left(\begin{array}{ccc|c}
1 & 1 & 0 & 2 \\
1 & 2 & 1 & 1 \\
1 & 3 & 2 & 0 \\
2 & 1 &-1 & 5
\end{array}\right)$$
I get the following:
$$\left(\begin{array}{ccc|c}
1 & 1 & 0 & 2 \\
1 & 2 & 1 & 1 \\
1 & 3 & 2 & 0 \\
2 & 1 &-1 & 5
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 0 & 2 \\
0 & 1 & 1 &-1 \\
0 & 2 & 2 &-2 \\
0 &-1 &-1 & 5
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 1 & 0 & 2 \\
0 & 1 & 1 &-1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
1 & 0 &-1 & 3 \\
0 & 1 & 1 &-1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}\right)
$$
Any of the last matrices can give you several solutions of the system. For example, $a=2$, $b=0$, $c=-1$. Or $a=3$, $b=-1$, $c=0$.
So you can see that $u$ is a linear combination of the vectors $v_1$, $v_2$, $v_3$ and it belongs to the span of these three vectors.
Let me try to explain another approach which also answers this question of yours:
If I were to find a basis for $U$ before this, can it be used to find if $u \in U$?
Let us start with a matrix which has $v_1$, $v_2$, $v_3$ and use row operations to transform it to reduced row echelon form.
$$
\left(\begin{array}{cccc}
1 & 1 & 1 & 2 \\
1 & 2 & 3 & 1 \\
0 & 1 & 2 &-1
\end{array}\right)\sim
\left(\begin{array}{cccc}
1 & 1 & 1 & 2 \\
0 & 1 & 2 &-1 \\
0 & 1 & 2 &-1
\end{array}\right)\sim
\left(\begin{array}{cccc}
1 & 1 & 1 & 2 \\
0 & 1 & 2 &-1 \\
0 & 0 & 0 & 0
\end{array}\right)\sim
\left(\begin{array}{cccc}
1 & 0 &-1 & 3 \\
0 & 1 & 2 &-1 \\
0 & 0 & 0 & 0
\end{array}\right)
$$
From this you see that the vectors $(1,0,-1,3)$ and $(0,1,2,-1)$ from the basis of $\operatorname{span}(v_1,v_2,v_3)$.
Now you could use the same approach as above; slight advantage is that now you only have two vectors.
But this is even easier. Since in the first two columns you have only one $1$ and the remaining entries are zeroes, you immediately see that to get a vector of the form $(2,1,\cdot,\cdot)$, the coefficients have to be $2$ and $1$.
So it only remains to check that $2\cdot(1,0,-1,3)+(0,1,2,-1)=(2,1,0,5)$.
Let me add also a third approach - kind of "guesswork". This will only work if there are not too many vectors and if they are simple, but occasionally it might be useful, since in this way sometimes you may be able to get the result faster.
For example, in this specific case, we have noticed that $v_2=v_1+v_3$. So this simplifies things a little -- we are now asking whether $u=(2,1,0,5)$ is a linear combination of $v_1=(1,1,1,2)$ and $v_3=(0,1,2,-1)$. So we want to express $u$ in the form $u=av_1+cv_3$ for some constants $a$, $c$.
Compare the first two coordinates. We may also notice that the first two coordinates of $v_1$ are equal. The same will be true for any multiple $av_1$. So the difference between these two coordinates in any vector of the form $x=av_1+cv_3$ will originate from $cv_3$. To be more precise, we will have $x_2-x_1=c(1-0)=c$.
So if we want to get vector $u$ with $u_2-u_1=-1$, then we necessarily have $c=-1$.
So we have
$$(2,1,0,5)=a(1,1,1,2)-(0,1,2,-1).$$
If we look at the first coordinate, then we have $2=a-0$, i.e., $a=2$ and we can check that
$$(2,1,0,5)=2(1,1,1,2)-(0,1,2,-1).$$
Some other coordinates. In the computation above we were "lucky" than two coordinates of one of the given vectors were equal. But basically the same trick works if we spot some simple relation between the coordinates, as long as it is linear. So let us try another example. We may notice that sum of the second and the fourth coordinate of $v_3$ is zero. So if we have $x=av_1+cv_3$, then $cv_3$ has zero contribution to the value $x_1+x_3$. Specifically, we have
$$x_1+x_3=a(1+2)=3a.$$
In this particular case we have $u_1+u_3=6$ and $a=2$, So we get
$$(2,1,0,5)=2(1,1,1,2)+c(0,1,2,-1).$$
Now it is relatively easy to calculate $c$ to get
$$(2,1,0,5)=2(1,1,1,2)-(0,1,2,-1).$$
