Is there NO solution to this linear system of 3 equations, $3$ unknowns? I have the following linear system: 
$$\begin{align}
&x + y  + 2z + 2 = 0 \\
&3x - y + 14z -6 = 0 \\
&x + 2y +5 = 0
\end{align}$$
I immediately noticed that there was no $z$ term in the last equation and thus determined that I will end with $0z = \text{some number}$ and therefore, came to the conclusion that there is no solution to the linear system. To see if I was right, I checked with echelon form and that also suggested that there was no solution. 
$$\begin{align}
&L_1: x + y  + 2z + 2 = 0 \\
&L_2: 3x - y + 14z -6 = 0 \\
&L_3: x + 2y +5 = 0 \\
\end{align}$$
Then $-3L_1 + L_2 \rightarrow L_2$
$$\begin{align}
&x + y + 2z + 2 = 0 \\
&0x + 3y - 42z - 18 = 0 \\
&x + 2y + 0z + 5 = 0 
\end{align}$$
Then $-L_1 + L_3 \rightarrow L_3$
$$\begin{align}
&x + y + 2z + 2 = 0 \\
&0x + 3y - 42z - 18 = 0 \\
&0x + 1y + 0z + 3 = 0 
\end{align}$$
Then $-L_2/3 + L_3 \rightarrow L_3$
$$\begin{align}
&x + y + 2z + 2 = 0 \\
&0x + 3y - 42z - 18 = 0 \\
&0x + 0y + 0z + 3 = 0 
\end{align}$$ 
Firstly, is my answer correct? If so, then can I make the same conclusions for other linear systems that are similar? 

Your first idea is off, because, for example the system $$x+y+z=1, x+2y=1, x+y+2z=1$$ still has a solution  even though $z$ is absent from the second equation.
On the other hand, your manipulation of equations (you call it echelon form) is one of possible correct methods of finding solutions/proving their absence.
