Does this vector lie in the vector subspace $U$? 
Given are the vectors $v_{1}=\begin{pmatrix} 1\\  3\\  5
\end{pmatrix}, v_{2}=\begin{pmatrix} 4\\  5\\  6 \end{pmatrix},
v_{3}=\begin{pmatrix} 6\\  4\\  2 \end{pmatrix}$ from
  $\mathbb{R}^{3}$.
The vector subspace $U = \text{span}\left\{v_{1},v_{2},v_{3}\right\}$
Does the vector $\begin{pmatrix} 2\\  4\\  5 \end{pmatrix}$ lie in
  $U$?

I'm not sure how this is done correctly but I think it's done by checking if the vector $\begin{pmatrix} 2\\  4\\  5 \end{pmatrix}$ is a linearly combination of $v_{1},v_{2},v_{3}$.
So I have written it all like that:
Let $a,b,c \in \mathbb{R}$.
$$a\begin{pmatrix}
1\\ 
3\\ 
5
\end{pmatrix}+b\begin{pmatrix}
4\\ 
5\\ 
6
\end{pmatrix}+c\begin{pmatrix}
6\\ 
4\\ 
2
\end{pmatrix}=\begin{pmatrix}
2\\ 
4\\ 
5
\end{pmatrix}$$
And then calculated each variable by using Gauss. In the end I had a wrong statement aka no solution and from this I conclude that the vector $\begin{pmatrix}
2\\ 
4\\ 
5
\end{pmatrix}$ doesn't lie in $U$.
I don't want write down all the calculation steps with Gauss because it would be too long, but is the way I did the correct way? Or is it done completely different?
 A: The complete matrix of the system, with Gaussian elimination:
\begin{align}
\begin{bmatrix}
1 & 4 & 6 & 2 \\
3 & 5 & 4 & 4 \\
5 & 6 & 2 & 5
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 4 & 6 & 2 \\
0 & -7 & -14 & -2 \\
0 & -14 & -28 & -5
\end{bmatrix}
&& \begin{aligned}R_2&\gets R_2-3R_1 \\ R_3&\gets R_3-5R_1\end{aligned}
\\&\to
\begin{bmatrix}
1 & 4 & 6 & 2 \\
0 & -7 & -14 & -2 \\
0 & 0 & 0 & -1
\end{bmatrix}
&& R_3\gets -2R_2
\end{align}
The system has no solution. I don't find this too long, do you?
Anyway, your method is correct, as far as I can see.
A: Equivalently, we want a+ 4b+ 6c= 2, 3a+ 5b+ 4c= 4, and 5a+ 6b+ 2c= 5.  Subtract the second equation from twice the third: 7a+ 7b= 6. Subtract the first equation from three times the third: 14a+ 14b= 13.  Multiplying 7a+ 7b= 6 by two, 14a+ 14b= 12.  a and b cannot satisfy both 14a+ 14b= 13 and 14a+ 14b= 12.  There are no numbers a and b that make the vector equation true.  $\begin{pmatrix} 2 \\ 4 \\ 5\end{pmatrix}$ cannot be written as a linear combination of $v_1$, $v_2$, and $v_3$ so is not in subspace U.
