Find all vectors in $\mathbb{R}^5$ such that $S$ is an orthogonal set. Express answer as a linear span. Find all vectors (x1, x2, x3, x4, x5) in  $\mathbb{R}^5$ such that S = {(1, 1, 1, 2, 0) , (0, 0, −2, 1, 0) , (0, 0, 0, 0, 1) , (x1, x2, x3, x4, x5)} is an orthogonal set. Express your answer as a linear span.
Clue was given to form a homogeneous system and solve it. I have tried solving
$$
  \left[\begin{array}{rrrrr|r}
    1 & 1 & 1 & 2 & 0 & 0\\
    0 & 0 & -2 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 & 1 & 0\\
  \end{array}\right]
$$
Attempt at a solution:
$$
x_2 = s  ,\,\,\, x_4 = t , \,\, where\,\, s,t \in \mathbb{R}, \\
x_1 = -s - \frac{5t}{2} , \, \, \, x_3 = \frac{t}{2}, \,\,\,x_5 = 0
\\\therefore\\(x_1,x_2,x_3,x_4,x_5) = (-s - \frac{5t}{2},\,\,\,s,\,\,\,\frac{t}{2},\,\,\,t,\,\,\,0)  
$$ 
But how do I answer in terms of linear span form here on?
Any help would be greatly appreciated !
EDIT:
With help from @Gerry's comments I got to
$$
(x_1,x_2,x_3,x_4,x_5) = span{\{(-1,1,0,0,0),(\frac{-5}{2},0, \frac{1}{2},1,0)\} }
$$
 A: My path to getting a solution.
We know that for orthogonal sets, any distinct pair has their dot product equals to zero
ie $ u_i.u_j = 0 \,\, for \,\,\,i \neq j $ 
The matrix way of doing that is simply 
$$
u_iu_j^T 
$$
Set up the system 
$$
  \left[\begin{array}{rrrrr}
    1 & 1 & 1 & 2 & 0 \\
    0 & 0 & -2 & 1 & 0 \\
    0 & 0 & 0 & 0 & 1 \\
  \end{array}\right]
  \left[\begin{array}{rrrrr}
    x_1 & x_2 & x_3 & x_4 & x_5 \\
  \end{array}\right]^T
  = 0
$$
Solve augmented matrix
$$
  \left[\begin{array}{rrrrr|r}
    1 & 1 & 1 & 2 & 0 & 0\\
    0 & 0 & -2 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 & 1 & 0\\
  \end{array}\right]
$$
Results
$$
x_2 = s  ,\,\,\, x_4 = t , \,\, where\,\, s,t \in \mathbb{R}, \\
x_1 = -s - \frac{5t}{2} , \, \, \, x_3 = \frac{t}{2}, \,\,\,x_5 = 0
\\\therefore\\(x_1,x_2,x_3,x_4,x_5) = (-s - \frac{5t}{2},\,\,\,s,\,\,\,\frac{t}{2},\,\,\,t,\,\,\,0)  
$$ 
Convert to linear span => Solution :
$$
(x_1,x_2,x_3,x_4,x_5) = span{\{(-1,1,0,0,0),(\frac{-5}{2},0, \frac{1}{2},1,0)\} }
$$
A: The vector $\begin{bmatrix}1&-1&0&0&0\end{bmatrix}$ is obviously orthogonal to all vectors given. Then using the adjugate matrix, as in this answer, we get a last orthogonal vector, $\begin{bmatrix}5&5&-2&-4&0\end{bmatrix}$.
Thus, an orthogonal basis for the subspace orthogonal to the three vectors given in the question is
$$
\begin{bmatrix}1&-1&0&0&0\\5&5&-2&-4&0\end{bmatrix}
$$
