Solution:

(i) Given that $(2 / x)+(3 / y)+(10 / z)=4$
$\begin{array}{l}
(4 / x)-(6 / y)+(5 / z)=1 \\
(6 / x)+(9 / y)-(20 / z)=2, x, y, z \neq 0
\end{array}$
We can write the given system in matrix form as:
$\begin{array}{l}
{\left[\begin{array}{ccc}
2 & 3 & 10 \\
4 & -6 & 5 \\
6 & 9 & -20
\end{array}\right]\left[\begin{array}{l}
\mathrm{u} \\
\mathrm{v} \\
\mathrm{w}
\end{array}\right]=\left[\begin{array}{l}
4 \\
1 \\
2
\end{array}\right]} \\
\mathrm{AX}=\mathrm{B}
\end{array}$
Now,
$\begin{array}{l}
|A|=2(75)-3(-110)+10(72) \\
=150+330+720 \\
=1200
\end{array}$
Therefore, the above system of equations has a unique solution, given by.
$X=A^{-1} B$
Cofactors of $\mathrm{A}$ are
$\begin{array}{l}
C_{11}=(-1)^{1+1} 120-45=75 \\
C_{21}=(-1)^{2+1}-60-90=150 \\
C_{31}=(-1)^{3+1} 15+60=75 \\
C_{12}=(-1)^{1+2}-80-30=110 \\
C_{22}=(-1)^{2+1}-40-60=-100 \\
C_{32}=(-1)^{3+1} 10-40=30 \\
C_{13}=(-1)^{1+2} 36+36=72 \\
C_{23}=(-1)^{2+1} 18-18=0 \\
C_{33}=(-1)^{3+1}-12-12=-24 \\
\text { Adj } A=\left[\begin{array}{ccc}
75 & 110 & 72 \\
150 & -100 & 0 \\
75 & 30 & -24
\end{array}\right]^{\mathrm{T}} \\
=\left[\begin{array}{ccc}
75 & 150 & 75 \\
110 & -100 & 30 \\
72 & 0 & -24
\end{array}\right]
\end{array}$
$A^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}$
Now, $X=A^{-1} B=\frac{1}{1200}\left[\begin{array}{ccc}75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24\end{array}\right]\left[\begin{array}{l}4 \\ 1 \\ 2\end{array}\right]$
$\begin{array}{l}
\mathrm{X}={ }^{1}{1200}\left[\begin{array}{l}
600 \\
400 \\
240
\end{array}\right] \\
{\left[\begin{array}{l}
\mathrm{u} \\
\mathrm{v} \\
\mathrm{W}
\end{array}\right]=\left[\begin{array}{l}
\frac{1}{2} \\
\frac{1}{3} \\
1 \\
\frac{1}{5}
\end{array}\right]}
\end{array}$
As a result, $X=2, Y=3$ and $Z=5$

(ii) Given that $x-y+2 z=7$
$\begin{array}{l}
3 x+4 y-5 z=-5 \\
2 x-y+3 z=12
\end{array}$
We can write the given system in matrix form as:
$\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{c}
7 \\
-5 \\
12
\end{array}\right]$
$A X=B$
Now,
$\begin{array}{l}
|A|=1(12-5)+1(9+10)+2(-3-8) \\
=7+19-22 \\
=4
\end{array}$
Therefore, the above system of equations has a unique solution, given by
$X=A^{-1} B$
Cofactors of $\mathrm{A}$ are
$\begin{array}{l}
C_{11}=(-1)^{1+1} 12-5=7 \\
C_{21}=(-1)^{2+1}-3+2=1 \\
C_{31}=(-1)^{3+1} 5-8=-3
\end{array}$
$\begin{array}{l}
\mathrm{C}_{12}=(-1)^{1+2} 9+10=-19 \\
\mathrm{C}_{22}=(-1)^{2+1} 3-4=-1 \\
\mathrm{C}_{32}=(-1)^{3+1}-5-6=11 \\
\mathrm{C}_{13}=(-1)^{1+2}-3-8=-11 \\
\mathrm{C}_{23}=(-1)^{2+1}-1+2=-1 \\
\mathrm{C}_{33}=(-1)^{3+1} 4+3=7
\end{array}$
$\operatorname{Adj} A=\left[\begin{array}{ccc}
7 & -19 & -11 \\
1 & -1 & -1 \\
-3 & 11 & -7
\end{array}\right]^{\mathrm{T}}$
$\begin{array}{l}
=\left[\begin{array}{ccc}
7 & 1 & -3 \\
-19 & -1 & 11 \\
-11 & -1 & 7
\end{array}\right] \\
A^{-1}=\frac{1}{|A|} \operatorname{adj} A
\end{array}$
Now, $X=A^{-1} B=\frac{1}{4}\left[\begin{array}{ccc}7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7\end{array}\right]\left[\begin{array}{c}7 \\ -5 \\ 12\end{array}\right]$
$\begin{array}{l}
X=\left[\begin{array}{c}
8 \\
4 \\
12
\end{array}\right] \\
{\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{y} \\
\mathrm{Z}
\end{array}\right]=\left[\begin{array}{l}
2 \\
1 \\
3
\end{array}\right]}
\end{array}$
As a result, $X=2, Y=1$ and $Z=3$