Linear Algebra vector matrix problem Link to the picture of the problem

So, do I simply put the numbers in a augmented matrix and into row echelon form, and solve for m? The variable m throws me off. 
 A: HINT: $-v_2+v_3$ will get the second coordinate of $w$ for you, but the first coordinate won't be right, so which multiple of $v_1$ would you need to add to fix it? Once you get the first two coordinates matched up, read off the third coordinate.
A: Your approach is correct. You want $w$ to be a linear combination of $v_1,v_2,v_3$ so we need:
$$
\left( \begin{matrix} 1&-5&4 \\ 0&2&-3 \\ 2&5&1 \end{matrix} \right) \left( \begin{matrix} a\\b\\c  \end{matrix} \right)= \left( \begin{matrix} 3\\-5\\m \end{matrix}\right)
$$
Row reducing the augmented matrix gives you this:
$$
\left(\begin{matrix} 1&0&0&\frac{7m-74}{31} \\ 0&1&0&\frac{3m+71}{31}\\ 0&0&1 & \frac{2m+63}{31} \end{matrix}\right)
$$
So you can make $m$ whatever you like and then the last column of the matrix above will give you the $a,b,c$ values that you need for the linear combination of $v_1,v_2,v_3$.
An easy way to see that any value of $m$ will do is to note that $v_1,v_2,v_3$ form a basis for $\mathbb{R}^3$ so any vector $w\in \mathbb{R}^3$ is a linear combination of them.
