Linear Algebra Question My question is; How can I find an equation relating $a,b$, and $c$ so that the linear system
$$\begin{cases}2x+y-z=a\\ x-2y-3z=b\\ -3x-y+2z=c\end{cases}$$
is consistent for any values of $a,b$, and $c$ that satisfy that equation.
Thanks,

I don't know what tools you have available, but here is how I would consider this problem.  If it uses terminology you do not know yet, you might come back to my answer later.  This is intentionally not a direct path to the solution, rather the path I went though to solve the problem.
Write this system as a matrix equation:
$$ 
\left( \begin{array}{ccc} 2 & 1 & -1 \\ 1 & -2 & -3 \\ -3 & -1 & 2 \end{array} \right)
\left( \begin{array}{c} x \\ y \\ z \end{array} \right)
= \left( \begin{array}{c} a \\ b \\ c \end{array} \right)
$$
If the matrix were invertible, we could multiply both sides by the inverse and get relations between $x,y,z$ and $a,b,c$.
$$
\left( \begin{array}{c} x \\ y \\ z \end{array} \right) =
\left( \begin{array}{ccc} 2 & 1 & -1 \\ 1 & -2 & -3 \\ -3 & -1 & 2 \end{array} \right)^{-1}
\left( \begin{array}{c} a \\ b \\ c \end{array} \right)
$$
But that matrix is not invertible: it has determinant 0, or you can see that the first column plus the third column equals the second column, so that the columns are not linearly independent.  There has to be a dependence among the rows as well, but it is not as obvious.  The "row reduced echelon form" is a good tool for finding column dependencies, so we use it on the transpose of the coefficient matrix (making the rows we want to relate into columns).
$$\text{rref} \left( \begin{array}{ccc} 2 & 1 & -3 \\ 1 & -2 & -1 \\ -1 & -3 & 2 \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & -7/5 \\ 0 & 1 & -1/5 \\ 0 & 0 & 0 \end{array} \right) $$
For solving $\alpha (2, 1, -1) + \beta(1,-2,-3) + \gamma(-3,-1,2) = 0$, this result means $\alpha - (7/5)\gamma = 0$ and $\beta - (1/5)\gamma = 0$.  Since $\gamma$ in independent, make things easy by choosing $\gamma = 5$, which gives $\alpha = 7$ and $\beta = 1$.
So on the left-hand side of the original system, $$7(2x+y-z) + (x-2y-3z) + 5(-3x-y+2z) = 0.$$  Keeping track of the right-hand side gives the dependency you want, $7a + b + 5c = 0$.  (It's interesting how row reduction and inner products as used by Luboš lead to the same result.)
