#### Answer

This system has unique solution and it is given by
$$x=1;\quad y= 0;\quad z= 3;\quad w=2.$$

#### Work Step by Step

To transform this system to Row-Echelon form we will follow the steps below
Step 1: Eliminate $x$ from the second equation by subtracting from it the first equation multiplied by $2$:
$$2x-2x + 3y-2y+0z-2z -w-2w = 0-12$$
which is then
$$y-2z-3w = -12.$$
Step 2: Eliminate $x$ from the 3rd equation by adding to it the first one multiplied by $3$:
$$-3x+3x+4y+3y+z+3z+2w+3w = 4+3\times6$$
which becomes
$$7y+4z+5w = 22$$
Step 3: Eliminate $x$ from the last equation by subtracting the first one from it:
$$x-x+2y-y-z-z+w-w = 0-6$$
which becomes
$$y-2z=-6.$$
Step 4: Now after we have done our eliminations we will just rewrite what we have:
\begin{align*}
x+y+z+w = &6\\
y-2z-3w=&-12\\
7y+4z+5w=&22\\
y-2z\qquad=&-6
\end{align*}
Step 5: Observe that we can directly get $w$ by subtracting the 2nd equation from the 4th:
$$y-y-2z-(-2z)+0w-(-3w) = -6-(-12)$$
which becomes $$3w = 6$$ and that is finally $$w=2.$$
Step 6: Put in the solution for $w$ in each equation and simplify it:
1st: $x+y+z+2=6\Rightarrow x+y+z = 4$
2nd: $y-2z-3\times 2 = -12 \Rightarrow y-2z= -6$
3rd: $7y+4z+5\times2 = 22\Rightarrow 7y+4z=12$
4th: $y-2z= -6$
Step 7: Rewrite the system by dropping the 4th equation since it is identical to the 2nd one:
\begin{align*}
x+y+z=&4\\
y-2z=&-6\\
7y+4z=&12
\end{align*}
Step 8: To eliminate $y$ from the 3rd equation substitute from it the second one multiplied by $7$:
$$7y-7y+4z-7\times(-2z)= 12-7\times(-6)$$
which becomes
$$18z= 54$$
and that gives
$$z = 3.$$
Step 9: Substitute the calculated value for $z$ into the second equation and find $y$:
$$y-2\times3 = -6$$
which becomes
$$y = 0.$$
Step 10: Substitute calculated values for $y$ and $z$ into the 1st equation and
find $x$:
$$x+0+3=4$$
which gives
$$x=1.$$
Step 11: Summarize all of the obtained results and write down the solution of the system:
$$x=1;\quad y= 0;\quad z= 3;\quad w=2.$$