converting a parametric R5 vector into a Cartesian form 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.

 A: 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 ${\cal R} (A)^\bot$.
