Solution:

(i) Given that $6 x-12 y+25 z=4$
$\begin{array}{l}
4 x+15 y-20 z=3 \\
2 x+18 y+15 z=10
\end{array}$
We can write the given system in matrix form as:
$\left[\begin{array}{ccc}6 & -12 & 25 \\ 4 & 15 & -20 \\ 2 & 18 & 15\end{array}\right]\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]=\left[\begin{array}{c}4 \\ 3 \\ 10\end{array}\right]$ Or $\mathrm{A} \mathrm{X}=\mathrm{B}$
$\mathrm{A}=\left[\begin{array}{ccc}6 & -12 & 25 \\ 4 & 15 & -20 \\ 2 & 18 & 15\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y} \\ \mathrm{z}\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{c}4 \\ 3 \\ 10\end{array}\right]$
Now, $|A|=6\left|\begin{array}{cc}15 & -20 \\ 18 & 15\end{array}\right|-12\left|\begin{array}{cc}4 & -20 \\ 2 & 15\end{array}\right|+25\left|\begin{array}{ll}4 & 15 \\ 2 & 18\end{array}\right|$
$=6(225+360)+12(60+40)+25(72-30)$
$=3510+1200+1050$
$=5760$
Therefore, the above system of equations has a unique solution, given by
$X=A^{-1} B$
Cofactors of $A$ are
$\begin{array}{l}
C_{11}=(-1)^{1+1}(225+360)=585 \\
C_{21}=(-1)^{2+1}(-180-450)=630 \\
C_{31}=(-1)^{3+1}(240-375)=-135 \\
C_{12}=(-1)^{1+2}(60+40)=-100 \\
C_{22}=(-1)^{2+1}(90-50)=40 \\
C_{32}=(-1)^{3+1}(-120-100)=220 \\
C_{13}=(-1)^{1+2}(72-30)=42 \\
C_{23}=(-1)^{2+1}(108+24)=-132 \\
C_{33}=(-1)^{3+1}(90+48)=138
\end{array}$
$\operatorname{Adj} A=\left[\begin{array}{ccc}585 & -100 & 42 \\ 630 & 40 & -132 \\ -135 & 220 & 138\end{array}\right]^{\mathrm{T}}$ $=\left[\begin{array}{ccc}585 & 630 & -135 \\ -100 & 40 & 220 \\ 42 & -132 & 138\end{array}\right]$
Now, $X=A^{-1} B=\frac{1}{5760}\left[\begin{array}{ccc}585 & 630 & -135 \\ -100 & 40 & 220 \\ 42 & -132 & 138\end{array}\right]\left[\begin{array}{c}4 \\ 3 \\ 10\end{array}\right]$
$X=\frac{1}{5760}\left[\begin{array}{l}2880 \\ 1920 \\ 1152\end{array}\right]$ $X=\left[\begin{array}{c}\frac{1}{2} \\ \frac{1}{7}\left[\frac{1}{3}\right. \\ \frac{1}{5}\end{array}\right]$ As a result, $X=\frac{1}{2}, Y=\frac{1}{3}$ and $Z=\frac{1}{5}$

(ii) Given that$3 x+4 y+7 z=14$
$\begin{array}{l}
2 x-y+3 z=4 \\
x+2 y-3 z=0
\end{array}$
We can write the given system in matrix form as:
$\begin{array}{l}
{\left[\begin{array}{ccc}
3 & 4 & 7 \\
2 & -1 & 3 \\
1 & 2 & -3
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{c}
14 \\
4 \\
0
\end{array}\right]_{\text {Or } \mathrm{A} X=\mathrm{B}} \\
{\mathrm{A}=\left[\begin{array}{ccc}
3 & 4 & 7 \\
2 & -1 & 3 \\
1 & 2 & -3
\end{array}\right], \mathrm{X}=\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{y} \\
\mathrm{Z}
\end{array}\right] \text { and } \mathrm{B}=\left[\begin{array}{c}
14 \\
4 \\
0
\end{array}\right]}}
\end{array}$
Now, $|\mathrm{A}|=3\left|\begin{array}{cc}-1 & 3 \\ 2 & -3\end{array}\right|-4\left|\begin{array}{cc}2 & 3 \\ 1 & -3\end{array}\right|+7\left|\begin{array}{cc}2 & 3 \\ 2 & -3\end{array}\right|$
$\begin{array}{l}
=3(3-6)-4(-6-3)+7(4+1) \\
=-9+36+35 \\
=62
\end{array}$
Therefore, the above system of equations has a unique solution, given by
$x=A^{-1} B$
Cofactors of $\mathrm{A}$ are
$\begin{array}{l}
C_{11}=(-1)^{1+1} 3-6=-3 \\
C_{21}=(-1)^{2+1}-12-14=26 \\
C_{31}=(-1)^{3+1} 12+7=19 \\
C_{12}=(-1)^{1+2}-6-3=9
\end{array}$
$C_{22}=(-1)^{2+1}-3-7=-10$
$C_{32}=(-1)^{3+1} 9-14=5$
$C_{13}=(-1)^{1+2} 4+1=5$
$C_{23}=(-1)^{2+1} 6-4=-2$
$C_{33}=(-1)^{3+1}-3-8=-11$
Adj $A=\left[\begin{array}{ccc}-3 & 9 & 5 \\ 26 & -5 & -2 \\ 19 & 5 & -11\end{array}\right]^{\mathrm{T}}$
$A^{-1}=\frac{1}{|A|} \operatorname{adj} A$
$X=\left[\begin{array}{ccc}-3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11\end{array}\right]$
$X=\left[\begin{array}{ccc}-3 & 26 & 19 \\ 9 & -16 & 5 \\ 5 & -2 & -11\end{array}\right]\left[\begin{array}{c}14 \\ 4 \\ 0\end{array}\right]$
$X=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$
As a result, $X=1, Y=1$ and $Z=1$