#### Answer

The solution of this system is unique and it is:
$$x= 1,\quad y=1,\quad z=-1.$$

#### Work Step by Step

As instructed in the problem we will put $A=1/x$, $B=1/y$ and $C=1/z$. This will give us the system of LINEAR equations in $A$, $B$ and $C$:
\begin{align*}
2A+B-2C=&5\\
3A-4B\qquad=&-1\\
2A+B+3C=&0.
\end{align*}
Now let us solve this system of equations.
Step 1: Subtract the first equation from the third one to eliminate $A$ and $B$:
$$2A-2A+B-B+3C-(-2C) = 0-5$$ which becomes $$5C=-5$$ and we get
$$C=-1.$$
Step 2: Put this calculated value of $C$ into the other two equations (where it appears) and make the system of two equations in $A$ and $B$:
$$2A+B-2(-1) = 5$$ and this becomes
$$2A+B+2=5$$ which is finally
$$2A+B=3.$$
There was no $C$ in the second equation so we will keep it as it is. Our system now reads:
\begin{align*}
2A+B=&3\\
3A-4B=&-1
\end{align*}
Step 3: Express $B$ in terms of $A$ from the first equation:
$$B=3-2A.$$
Step 4: Put this into the second equation and solve for $A$:
$$3A-4\times(3-2A) = -1\Rightarrow 3A-12+8A = -1\Rightarrow 11A = 11$$
and this gives us
$$A=1.$$
Step 5: Use this calculated value to find $B$:
$$B= 3-2A = 3-2 = 1.$$
Now let's find $x$, $y$ and $z$.
Since $A=1/x$ then $$x=\frac{1}{A} = 1.$$
Since $B=1/y$ then $$y=\frac{1}{B} = 1$$
Since $C=1/z$ then $$z=\frac{1}{C} = \frac{1}{-1} = -1$$