Is this linearly independent? What is the dimension span? 
Consider the vectors in $\mathbb{R}^4$ defined by
  $v_1 = (1,2,10,5), v_2 = (0,1,1,1), v_3 = (1,4,12,7)$.
  Find the reduced row echelon form of the matrix which has these as its rows. What is its rank? Is $\{v_1 , v_2 , v_3 \}$ linearly independent? What is the dimension of $ \text{span}(v_1, v_2, v_3)$?
  I'm not sure if i'm right.

I was able to take the system $av_1+av_2+av_3=0$
$$\begin{bmatrix} 1&0&1\\2&1&4\\10&1&12\\5&1&7 \end{bmatrix} = \begin{bmatrix} 0\\0\\0\\0 \end{bmatrix}$$
And reduce it down to 
$$\begin{bmatrix} 1&0&1\\0&1&2\\0&0&0\\0&0&0\end{bmatrix}= \begin{bmatrix} 0\\0\\0\\0 \end{bmatrix}$$
which means rank is $2$ because we have $2$ leading ones.
Also i think it's not linearly independent as row 1 $= a+c=0, \text{row} \hspace{.1cm} 2 = b+2c=0, a=-c$ and $b=-2c$
hence $a$,$b$ and $c$ are not equal to zero.
If it was linearly independent this would mean that the basis are vectors 1 and 2 but not 3. Meaning $v_1$ and $v_2$  spans $\mathbb{R}^2$ and hence the dimension span is $\mathbb{R}^4$
There is also a second part which i don't understand. But it's ok i'm mainly interested in the previous part might post this on a separate question.

Find a homogeneous linear system for which the space of solutions is exactly the subspace of $\mathbb{R}^4$ spanned by
  $v_1, v_2,$ and $v_3$.


The vectors are linearly dependent since $v_3=v_1+2v_2$. As you noted, you have the two leading ones, and it is clear that the vectors span a space with dimension two, the say way that $Span\{\vec i,\vec j\}$ spans two dimensions in $\Bbb R^3$
