What are the respective different ranks of the matrix ?
I tried with all parameters $a,b$ and $c$ being zero , and then $c$ being $0$. $$\left( \begin{array}{ccc} 1 & 4 & 3 \\ 5 & a & b \\ c & 8 & 6 \end{array} \right)$$ Is this OK to get all possible solutions depending on and parameters $a,b$ and $c$, which are all real numbers?
Applying Gauss-Jordan
$$\left( \begin{array}{ccc} 1 & 4 & 3 \\ 5 & a & b \\ c & 8 & 6 \end{array} \right)$$
$$\left( \begin{array}{ccc} 1 & 4 & 3 \\ 0 & a-20 & b-15 \\ 0 & 8-4c & 6-3c \end{array} \right)$$
Case 1: If $a\neq20$ then
$$\left( \begin{array}{ccc} 1 & 4 & 3 \\ 0 & 1 & \frac{b-15}{a-20} \\ 0 & 8-4c & 6-3c \end{array} \right)$$
$$\left( \begin{array}{ccc} 1 & 4 & 3 \\ 0 & 1 & \frac{b-15}{a-20} \\ 0 & 8-4c & 6-3c \end{array} \right)$$
$$\left( \begin{array}{ccc} 1 & 0 & 3-\frac{4(b-15)}{a-20} \\ 0 & 1 & \frac{b-15}{a-20} \\ 0 & 0 & 6-3c-\frac{4(2-c)(b-15)}{a-20} \end{array} \right)$$
$$\left( \begin{array}{ccc} 1 & 0 & \frac{3a-4b}{a-20} \\ 0 & 1 & \frac{b-15}{a-20} \\ 0 & 0 & \frac{(2-c)(3a-4b)}{a-20} \end{array} \right)$$
If $c\neq2$ and $3a\neq4b$ then all are independent and you have rank 3.
If $c=2$ or $3a=4b$ then you have rank 2.
Case 2: If $a=20$ then
$$\left( \begin{array}{ccc} 1 & 4 & 3 \\ 0 & 0 & b-15 \\ 0 & 8-4c & 6-3c \end{array} \right)$$
$$\left( \begin{array}{ccc} 1 & 4 & 3 \\ 0 & 8-4c & 6-3c \\ 0 & 0 & b-15 \end{array} \right)$$
If $c=2$ and $b=15$ then rank 1. If $c=2$ or $b=15$ then rank 2.
Edit:
Looking back at the second line we could have worked everything out from there:
$$\left( \begin{array}{ccc} 1 & 4 & 3 \\ 0 & a-20 & b-15 \\ 0 & 8-4c & 6-3c \end{array} \right)$$
If $a\neq20$ and $b\neq15$ and $c\neq2$ then rank 3. If $c=2$ OR $a=20$ & $b=15$ then rank 2. If $c=2$ and $a=20$ and $b=15$ then rank 1.