Solution:

(i) Given that$5 x+3 y+7 z=4$
$\begin{array}{l}
3 x+26 y+2 z=9 \\
7 x+2 y+10 z=5
\end{array}$
We can write this as:
$\begin{array}{l}
{\left[\begin{array}{ccc}
5 & 3 & 7 \\
3 & 26 & 2 \\
7 & 2 & 10
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{l}
4 \\
9 \\
5
\end{array}\right]} \\
|\mathrm{A}|=5(260-4)-3(30-14)+7(6-182) \\
=5(256)-3(16)+7(176) \\
|\mathrm{A}|=0
\end{array}$
Therefore, $\mathrm{A}$ is singular. Hence, the given system is either inconsistent or it is consistent with infinitely many solution according to as:
$(\operatorname{Adj} A) \times B \neq 0$ or $(A d j A) \times B=0$
Cofactors of $\mathrm{A}$ are
$\begin{array}{l}
C_{11}=(-1)^{1+1} 260-4=256 \\
C_{21}=(-1)^{2+1} 30-14=-16 \\
C_{31}=(-1)^{3+1} 6-182=-176 \\
C_{12}=(-1)^{1+2} 30-14=-16 \\
C_{22}=(-1)^{2+1} 50-49=1 \\
C_{32}=(-1)^{3+1} 10-21=11 \\
C_{13}=(-1)^{1+2} 6-182=-176 \\
C_{23}=(-1)^{2+1} 10-21=11 \\
C_{33}=(-1)^{3+1} 130-9=121 \\
\quad \operatorname{Adj} A=\left[\begin{array}{ccc}
256 & -16 & -176 \\
-16 & 1 & 11 \\
-176 & 11 & 121
\end{array}\right]^{\mathrm{T}} \\
=\left[\begin{array}{ccc}
256 & -16 & -176 \\
-16 & 1 & 11 \\
-176 & 11 & 121
\end{array}\right]
\end{array}$
$\operatorname{Adj} A \times B=\left[\begin{array}{ccc}256 & -16 & -176 \\ -16 & 1 & 11 \\ -176 & 11 & 121\end{array}\right]\left[\begin{array}{l}4 \\ 9 \\ 5\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
Now, AX = B has infinite many solution
Suppose $z=k$
Therefore, $5 x+3 y=4-7 k$
$3 x+26 y=9-2 k$
We can write this as
$\begin{array}{l}
{\left[\begin{array}{cc}
5 & 3 \\
3 & 26
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y}
\end{array}\right]=\left[\begin{array}{l}
4-7 \mathrm{k} \\
9-2 \mathrm{k}
\end{array}\right]} \\
|\mathrm{A}|=121
\end{array}$
$\operatorname{Adj} A=\left[\begin{array}{cc}26 & -3 \\ -3 & 5\end{array}\right]$
Now, $X=A^{-1} B=\frac{1}{|A|}$ Adj $A \times B$
$\begin{array}{l}
=\frac{1}{121}\left[\begin{array}{cc}
26 & -3 \\
-3 & 5
\end{array}\right]\left[\begin{array}{l}
4-7 \mathrm{k} \\
9-2 \mathrm{k}
\end{array}\right] \\
=\frac{1}{121}\left[\begin{array}{l}
77-176 \mathrm{k} \\
11 \mathrm{k}+33
\end{array}\right] \\
{\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{c}
\frac{7-16 \mathrm{k}}{11} \\
\frac{\mathrm{k}+3}{11}
\end{array}\right]}
\end{array}$
There values of $x, y$ and $z$ satisfy the third equation

(ii) Given that $x+y+z=6$
$\begin{array}{l}
x+2 y+3 z=14 \\
x+4 y+7 z=30
\end{array}$
We can write this as:
$\begin{array}{l}
{\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 3 \\
1 & 4 & 7
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y} \\
\mathrm{z}
\end{array}\right]=\left[\begin{array}{c}
6 \\
14 \\
30
\end{array}\right]} \\
|\mathrm{A}|=1(2)-1(4)+1(2)
\end{array}$
$\begin{array}{l}
=2-4+2 \\
|A|=0
\end{array}$
Therefore, A is singular. Hence, the given system is either inconsistent or it is consistent with infinitely many solution according to as:
$(\operatorname{Adj} A) \times B \neq 0$ or $(\operatorname{Adj} A) \times B=0$
Cofactors of A are
$\begin{array}{l}
\mathrm{C}_{11}=(-1)^{1+1} 14-12=2 \\
\mathrm{C}_{21}=(-1)^{2+1} 7-4=-3 \\
C_{31}=(-1)^{3+1} 3-2=1 \\
C_{12}=(-1)^{1+2} 7-3=-4 \\
C_{22}=(-1)^{2+1} 7-1=6 \\
C_{32}=(-1)^{3+1} 3-1=2 \\
C_{13}=(-1)^{1+2} 4-2=2 \\
C_{23}=(-1)^{2+1} 4-1=-3 \\
C_{33}=(-1)^{3+1} 2-1=1
\end{array}$
$\operatorname{Adj} A=\left[\begin{array}{ccc}2 & -4 & 2 \\ -3 & 6 & -3 \\ 1 & -2 & 1\end{array}\right]^{\mathrm{T}}$
$=\left[\begin{array}{ccc}
2 & -3 & 1 \\
-4 & 1 & -2 \\
2 & -3 & 1
\end{array}\right]$
Adj $A \times B=\left[\begin{array}{ccc}2 & -3 & 1 \\ -4 & 1 & -2 \\ 2 & -3 & 1\end{array}\right]\left[\begin{array}{c}6 \\ 14 \\ 30\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
Now, $A X=B$ has infinite many solution
Suppose $z=k$
Therefore, $x+y=6-k$
$x+2 y=14-3 k$
We can write this as:
$\left[\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y}
\end{array}\right]=\left[\begin{array}{c}
6-\mathrm{k} \\
14-3 \mathrm{k}
\end{array}\right]$
$\begin{array}{l}
|\mathrm{A}|=1 \\
\text { Adj } \mathrm{A}=\left[\begin{array}{cc}
2 & -1 \\
-1 & 1
\end{array}\right]
\end{array}$
Now, $X=A^{-1} B=\frac{1}{|A|}$ Adj $A \times B$
$\begin{array}{r}
=\frac{1}{1}\left[\begin{array}{cc}
2 & -1 \\
-1 & 1
\end{array}\right]\left[\begin{array}{c}
6-k \\
14-3 k
\end{array}\right] \\
=\frac{1}{1}\left[\begin{array}{l}
12-2 \mathrm{k}-14+3 \mathrm{k} \\
-6+\mathrm{k}+14-3 \mathrm{k}
\end{array}\right] \\
{\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{c}
-2+k \\
8-2 k
\end{array}\right]}
\end{array}$
There values of $x, y$ and z satisfy the third equation
As a result, $x=k-2, y=8-2 k, z=k$