<s>[INST] Matrixes with common parameters to result in no inverse I've been given three matrices $A, B \ \& \ C$ which are defined as follows:
$$ A = {
\left[
\begin{array}{ccc}
b & 5 & 8 \\ 
c & 1 & 3 \\ 
a & 4 & 3 \\
\end{array}
\right]
},\ B = {
\left[
\begin{array}{ccc}
7 & 4 & 2 \\ 
5 & 5 & -1 \\ 
-2 & -a & -b \\
\end{array}
\right]
}, \ C = {
\left[
\begin{array}{ccc}
3 & c & 5 \\ 
1 & b & 4 \\ 
2 & a & 1 \\
\end{array}
\right]
} $$
First, I've been asked to find the determinant of each matrix in terms of the parameters, and I've done it like this:
$ |A| = b \begin{vmatrix} 
1 & 3 \\ 
4 & 3 \\ 
\end{vmatrix} - 5 \begin{vmatrix} 
c & 3 \\ 
a & 3 \\ 
\end{vmatrix} + 8 \begin{vmatrix} 
c & 1 \\ 
a & 4 \\ 
\end{vmatrix} = \ b(3-12)-5(3c-3a)+8(4c-a) = \\ -9b - 15c + 15a + 32c - 8a = \mathbf{7a - 9b + 17c} $
$ |B| = 7 \begin{vmatrix} 
5 & -1 \\ 
-a & -b \\ 
\end{vmatrix} - 4 \begin{vmatrix} 
5 & -1 \\ 
-2 & -b \\ 
\end{vmatrix} + 2 \begin{vmatrix} 
5 & 5 \\ 
-2 & -a \\ 
\end{vmatrix} = 7(-5b-a)-4(-5b-2)+2(-5a+(-10)) = -35b - 7a + 20b + 8 - 10a + 20 = \mathbf{-17a -15b + 28} $
$ |C| = 3 \begin{vmatrix} 
b & 4 \\ 
a & 1 \\ 
\end{vmatrix} - c \begin{vmatrix} 
1 & 4 \\ 
2 & 1 \\ 
\end{vmatrix} + 5 \begin{vmatrix} 
1 & b \\ 
2 & a \\ 
\end{vmatrix} = \ 3(b - 4a) - c(1 - 8) + 5(a - 2b) = \\ 3b - 12a – c + 8c + 5a - 10b = \mathbf{-7a -7b + 7c} $
So after this, I have to calculate the value of the parameters to make all three of the matrices not invertible, so I assume $|M| = 0$ has to be true for each matrix. But I don't know how to proceed.
I thought of defining a system of linear equations using the three of them equaled to $ 0 $ but as $|B|$ has no $c$ term I don't know how to acomplish this.
 [/INST] Your determinants are right. Next you set them equal to $0$. You do not ignore the number 28. You put it on the RHS as mentioned in the comments. 
$$\left(
\begin{array}{ccc|c}
7 & -9 & 17 & 0\\
-17 & -15 & 0 & -28\\
-7 & -7 & 7 &0 \\
\end{array} \right)$$
Now you can use the Gaussian elimination algorithm to obtain the solution. Here are the two steps.
Step 1
Multiply row 1 by $17/7$ and add to row 2. 
Add row 1 to row 3.
$$\left(
\begin{array}{ccc|c}
7 & -9 & 17 & 0\\
0 & -258/7  & 289/7  & -28\\
0 & -16  & 24 &0 \\
\end{array} \right)$$
Step 2
Multiply row 2 by $-56/129$ and add to row 3.
$$\left(
\begin{array}{ccc|c}
7 & -9 & 17 & 0\\
0 & -258/7  & 289/7  & -28\\
0 & 0 & 784/129 &1568/129 \\
\end{array} \right)$$
</s>