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 $v_3$.

 A: 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 $\Bbb R^3$
A: First, you make the vectors into the columns of the matrix, not its rows...I'm assuming that's a typo. While most of what you write is morally correct, some of it is written incorrectly in nonsensically. First, you've written the equation $A=0$, where $A$ is your matrix. This is not a system, and it's also not a valid equation, because $A$ is not the zero matrix. What you should write is $Ax=0$, if you want a system, and this represents the vector equation $x_1 v_1+x_2v_2+x_3v_3=0$, not $av_1+av_2+av_3=0$, i.e., you need to allow distinct scalars for each of the vectors $v_i$. 
I'll assume your row reduction is correct. Then yes, as there are two leading $1$'s in the reduced row echelon form of $A$, and the corresponding  columns of the original matrix form a basis for the column space of $A$, the rank of $A$ (i.e. the dimension of its column space) is equal to $2$. 
I don't understand what you've written about linear independence involving rows because you haven't introduced the notation. Linear dependence of the columns of your matrix $A$ is indicated by the presence of free columns in the reduced row echelon form, also known as non-pivot columns, which are columns that don't possess a leading $1$. There is one of these in your reduced row echelon form, namely the third column, and this implies the columns of the original matrix are linearly dependent. Indeed, your reduced row echelon form allows you to write down all solutions to the system $Ax=0$, which, when viewed as the vector equation above, describes all linear dependence relations among the columns of $A$. You find that $v_1+2v_2-v_3=0$, a non-trivial dependence relation. 
The sentence beginning "If it was..." doesn't really make sense as written, but I believe what you're intending is that the first two columns, $v_1$ and $v_2$, form a basis for the column space of $A$, which is true. The pivot columns of the original matrix always form a basis for the column space of the matrix. They don't span $\mathbf{R}^2$, and ``dimension span is $\mathbf{R}^4$" doesn't make sense. They span a $2$-dimensional subspace of $\mathbf{R}^4$. 
A: $v_1, v_2$ and $v_3$ are not  linearly independent, because the row-rank of a matrix is equal to its column-rank. Hence the system of vectors has rank $2$.
For the second part (finding a linear relation between $v_1, v_2$ and $v_3 $) this means solving the linear system $A\mathbf x=\mathbf 0$, where $A$ is the matrix with column-vectors $v_1, v_2, v_3 $. As you've written the matrix in row-reduced echelon form, the solutions are at hand:
$$x=-z,\quad y=-2z$$
Set $z=-1$, and you have the relation:
$$v_1+2v_2-v_3=0.$$
Added: I misunderstood the question in the second part. What has to be found is a set of linear equations that define the subspace generated by $v_1, v_2$  and $v_3$. As this subspace has dimension $2$ it has codimension $\dim\mathbf R^4-2=2$, i.e. we must find a set of $2$ linearly independent equations that are satisfied by $v_1,v_2$ and $v_3$. Let $ax+by+cz+dt=0$ any such equation. Write it must be satisfied by the $3$vectors:
$$\begin{bmatrix}a&b&c&d\end{bmatrix}\begin{bmatrix} 1&0&1\\2&1&4\\10&1&12\\5&1&7 \end{bmatrix}= \begin{bmatrix} a+2b+10c+5d\\b+c+d\\a+4b+12c+7d\end{bmatrix}= \begin{bmatrix}1&2&10&5\\0&1&1&1\\1&4&12&7\end{bmatrix}\begin{bmatrix} a\\b\\c\\d\end{bmatrix}=\begin{bmatrix} 0\\0\\0\end{bmatrix}$$
We see we have to row-reduce the transposed matrix, which is readily done:
$$\begin{bmatrix}1&0&8&3\\0&1&1&1\\0&0&0&0\end{bmatrix}$$
It indeed has rank $2$ (its rank is the codimension of the system of vectors) and the solutions are given by:
$$a=-8c-3d,\quad b=-c-d.$$
To have two linearly indpendent equations, we'll take successively: $c=-1,d=0$  and $c=0,d=-1$, whence the equations:
$$8x+y-z=0,\quad 3x+y-t=0.$$
