let $$A_c = \left( \begin{array}{ccc} 2+2c&-4+2c&-4-3c+c^2\\ 1+c&-1+c&-1-c+c^2\\ -1-c&2-c&2+c\\ \end{array} \right) $$
$$ b_c= \left( \begin{array}{ccc|c} 11-5c\\ 3-2c\\ -6+3c\\ \end{array} \right) $$
How can I define $c_1,c_2,c_3 \in \mathbb R$ with $c_1 \lt c_2$ so, that the linear equationsystems $(A_{c_1}|b_{C_1})$ and $(A_{c_2}|b_{C_2})$ so that they have no solution and also $(A_{c_3}|b_{C_3})$ that is has infinite solutions?
$$ \left( \begin{array}{ccc|c} 2+2c&-4+2c&-4-3c+c^2&11-5c\\ 1+c&-1+c&-1-c+c^2&3-2c\\ -1-c&2-c&2+c&-6+3c\\ \end{array} \right) $$
I know that for infinite solutions two lines have to be equal:
$(1+c)x_1 + (-1+c)x_2+(-1-c+c^2)x_3-3+2c=(-1-c)x_1+(2-c)x_2+(2+c)x_3+6-3c$
$\equiv (2+c)x_1 + (-3+2c)x_2 + (-3-2c+c^2)x_3-9+5c=0$
But how to solve that?
And how can I get the $c_1, c_2$ for the matrices without any solution?
EDIT:
I have simplyfied the matrix to $$ \left( \begin{array}{ccc|c} 2+2c&-4+2c&-4-3c+c^2&11-5c\\ 0&-2&-2-c-c^2&5-c\\ 0&0&-c+c^2&-1+c\\ \end{array} \right) $$
so the $c_1, c_2$ for no solution are $-c+c^2=0 \Rightarrow c=0, c=1$
How can I get $c_3$?