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 $v_3$.


$v_1, v_2$ 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:
$$8x+y-z=0,\quad 3x+y-t=0.$$
