Find a homogeneous system whose solution space is spanned by the given vectors find a homogeneous system whose solution space is spanned by the following set of 3 vectors: 
$$(1,-2,0,3,-1) , (2,-3,2,5,-3), (1,-2,1,2,-2)$$
Please help, I've only seen similar questions where there are 4 unknowns not 5

Let
$$
v_1 = (1, -2, 0, 3, -1),\\
v_2 = (2, -3, 2, 5, -3),\\
v_3 = (1, -2, 1, 2, -2).
$$
and $L = \mathop{\mathrm{span}}(v_1, v_2, v_3)$. Let's find $L^\perp = \{x\mid \forall v\in L\colon (v, x) = 0\}$:
$$
L^\perp\colon\left\{\begin{aligned}
x_1 -2x_2 + 3x_4 -x_5&=0,\\
2x_1 -3x_2 + 2x_3 + 5x_4 -3x_5&=0,\\
x_1 -2x_2 + x_3 + 2x_4 -2x_5&=0.
\end{aligned}\right.
$$
Solve this system:
$$
\begin{aligned}
x_1 &= t_1,\\
x_2 &= t_1 + 4t_2,\\
x_3 &= -t_1-4t_2,\\
x_4 &= t_2,\\
x_5 &= -t_1-5t_2,
\end{aligned}
$$
or
$$
x = \begin{pmatrix}1\\1\\-1\\0\\-1\end{pmatrix} t_1 +
\begin{pmatrix}0\\4\\-4\\1\\-5\end{pmatrix} t_2 = u_1t_1 + u_2 t_2.
$$
But $L = (L^\perp)^\perp$. Do the same trick and
$$
L\colon\left\{\begin{aligned}
x_1 + x_2 -x_3  - x_5 &= 0\\
4x_2-4x_3+x_4-5x_5 &= 0
\end{aligned}\right.
$$
is needed system.
