Solution:

(i) Given that $3 x-y-2 z=2$
$\begin{array}{l}
2 y-z=-1 \\
3 x-5 y=3 \\
{\left[\begin{array}{ccc}
3 & -1 & -2 \\
0 & 2 & -1 \\
3 & -5 & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
2 \\
-1 \\
3
\end{array}\right]} \\
|A|=3(-5)+1(3)-2(-6) \\
|A|=0
\end{array}$
Cofactors of A are
$\begin{array}{l}
C_{11}=(-1)^{1+1} 0-5=-5 \\
C_{21}=(-1)^{2+1} 0-10=10 \\
C_{31}=(-1)^{3+1} 1+4=5 \\
C_{12}=(-1)^{1+2} 0+3=-3 \\
C_{22}=(-1)^{2+1} 0+6=6 \\
C_{32}=(-1)^{3+1}-3-0=3 \\
C_{13}=(-1)^{1+2} 0-6=-6 \\
C_{23}=(-1)^{2+1}-15+3=12 \\
C_{33}=(-1)^{3+1} 6-0=6
\end{array}$
$\begin{array}{l}
\text { Adj } A=\left[\begin{array}{ccc}
-5 & 3 & -6 \\
10 & 6 & 12 \\
5 & 3 & 6
\end{array}\right]^{\mathrm{T}} \\
=\left[\begin{array}{ccc}
-5 & 10 & 5 \\
3 & 6 & 3 \\
-6 & 12 & 6
\end{array}\right] \\
\text { Adj } A \times B=\left[\begin{array}{ccc}
-5 & 10 & 5 \\
3 & 6 & 3 \\
-6 & 12 & 6
\end{array}\right]\left[\begin{array}{c}
2 \\
-1 \\
3
\end{array}\right] \\
{\left[\begin{array}{c}
-10-10+15 \\
6-6+9 \\
-12-12+18
\end{array}\right] \neq\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]}
\end{array}$
As a result, the above system is inconsistent.

(ii) Given that $x+y-2 z=5$
$\begin{array}{l}
x-2 y+z=-2 \\
-2 x+y+z=4
\end{array}$
$\begin{array}{l}
{\left[\begin{array}{ccc}
1 & 1 & -2 \\
1 & -2 & 1 \\
-2 & 1 & 1
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{c}
5 \\
-2 \\
4
\end{array}\right]} \\
|\mathrm{A}|=1(-3)-1(3)-2(-3)=-3-3+6 \\
|\mathrm{~A}|=0
\end{array}$
Cofactors of A are:
$\begin{array}{l}
C_{11}=(-1)^{1+1}-2-1=-3 \\
C_{21}=(-1)^{2+1} 1+2=-3 \\
C_{31}=(-1)^{3+1} 1-4=-3 \\
C_{12}=(-1)^{1+2} 1+2=-3 \\
C_{22}=(-1)^{2+1} 1-4=-3 \\
C_{32}=(-1)^{3+1} 1+2=-3 \\
C_{13}=(-1)^{1+2} 1-4=-3 \\
C_{23}=(-1)^{2+1} 1+2=-3 \\
C_{33}=(-1)^{3+1}-2-1=-3
\end{array}$
$\begin{array}{l}
\text { Adj } A=\left[\begin{array}{rrr}
-3 & -3 & -3 \\
-3 & -3 & -3 \\
-3 & -3 & -3
\end{array}\right]^{\mathrm{T}} \\
=\left[\begin{array}{lll}
-3 & -3 & -3 \\
-3 & -3 & -3 \\
-3 & -3 & -3
\end{array}\right]
\end{array}$
$\begin{array}{l}
\text { Adj } A \times B=\left[\begin{array}{lll}
-3 & -3 & -3 \\
-3 & -3 & -3 \\
-3 & -3 & -3
\end{array}\right]\left[\begin{array}{c}
5 \\
-2 \\
4
\end{array}\right] \\
=\left[\begin{array}{l}
-15+6-12 \\
-15+6-12 \\
-15+6-12
\end{array}\right] \neq\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]
\end{array}$
As a result, the above system is inconsistent.