How do you solve a problem like this. I'm completely stumped. it seems like there should be an easy solution but I'm obviously over looking it. any help would be greatly appreciated.

How do you solve a problem like this. I'm completely stumped. it seems like there should be an easy solution but I'm obviously over looking it. any help would be greatly appreciated.

Let $x_0 = (1, 0,2,1,0)^T$, $a_0 = (-2, -1, 2, -1 , 1)^T$, $a_1=(0,2,-2,1,-1)^T$, $a_2=(-4,0,2,-1,1)^T$.
Let $\cal P$ denote the set in question, ie, ${\cal P} = \{ x_0+\sum_{k=0}^2 \lambda_k a_k \}_{\lambda \in \mathbb{R}^3}$. A quick check shows that $a_0 = \frac{1}{2} (a_2-a_1)$, so in fact ${\cal P} = \{ x_0+\sum_{k=1}^2 \lambda_k a_k \}_{\lambda \in \mathbb{R}^2}$, and $a_1, a_2$ are linearly independent. Let $A=\begin{bmatrix} a_1 & a_2 \end{bmatrix}$. Then, with a slight abuse of notation, we can write ${\cal P} = x_0+{\cal R} (A)$, where ${\cal R} (A)$ denotes the range of $A$.
The goal is to find a matrix $M$ such that ${\cal P} = \{ x | M(x-x_0) = 0\} = \{ v+x_0\}_{v \in \ker M}$. Equivalently, we want to find $M$ such that ${\cal P}-x_0 = {\cal R} (A) = \ker M$. (The point being that $x_0$ is sort of irrelevant here.)
We note that $x \in {\cal R} (A) $ iff $x \bot {\cal R} (A)^\bot$, so if we can find a basis $c_1,...,c_k$ for ${\cal R} (A)^\bot$, then letting $C = \begin{bmatrix} c_1 & \cdots & c_k \end{bmatrix}$, we have $x \in {\cal R} (A) $ iff $C^T x = 0$. (Then letting $M=C^T$ finishes the problem.) We note in passing that since $\dim {\cal R} (A) = 2$, we have $\dim {\cal R} (A)^\bot = 3$.
There are many ways to find a basis of ${\cal R} (A)^\bot$. Tedious inspection yields $$C=\begin{bmatrix} 0 & 1 & -2 \\ 0 & 2 & -4 \\ 0 & 1 & -18 \\ 1 & -1 & -14 \\ 1 & 1 & 14 \end{bmatrix}$$
From this we obtain the desired description $\cal P$ is the set of $(x_1,...,x_5) \in \mathbb{R}^5$ that satisfy: \begin{eqnarray} (x_4-1)+x_5 & = & 0 \\ (x_1-1) + 2 x_2 + (x_3-2) - (x_4-1) + x_5 & = & 0 \\ -2(x_1-1) -4 x_2 -18(x_3-2) - 14(x_4-1) + 14 x_5 & = & 0 \\ \end{eqnarray}
A more computational approach would be to compute matrix $\Pi$ of the orthogonal projection onto ${\cal R} (A)^\bot$, and select a maximal set of linearly independent columns of $\Pi$. From the least squares problem, it is straightforward to show that $\Pi = A(A^TA)^{-1} A^T - I$. Another tedious computation shows that $$46 \, \Pi =\begin{bmatrix} 6 & 12 & 8 & -4 & 4 \\ 12 & 24 & 16 & -8 & 8 \\ 8 & 16 & 26 & 10 & -10 \\ -4 & -8 & 10 & 41 & 5 \\ 4 & 8 & -10 & 5 & 41 \end{bmatrix}$$ It is straightforward to verify that columns $2,4$ of $\Pi$ are multiples of columns $1,5$, hence columns $1,3,5$ provide another basis for ${\cal R} (A)^\bot$.