Let there be: $$2\cdot a\cdot y+\left(a+4\right)\cdot x = 0$$
$$\left(2\cdot a+6\right)\cdot z+\left(16\cdot a-16\right)\cdot y+ \left(6\cdot a+24\right)\cdot x = 4\cdot a-2$$
$$\left(8\cdot a+16\right)\cdot z+\left(54\cdot a-56\right)\cdot y+ \left(20\cdot a+80\right)\cdot x = 14\cdot a-8$$
For which $a$ there is no solution/one solution/infinite solutions
The RREF I got is: $$\ \left[ \begin{array}{ccc|c} a+4 & 2a & 0 & 0 \\ 0 & 2a-8 & a+3 & 2a-1 \\ 0 & 0 & a-5 & -1 \\ \end{array} \right] $$
definitely $a=5$ ,How should I proceed from here for one solution/infinite solutions? (when I need to provide 3 $a$ for one solution and 2 $a$ for infinite solutions)