Solution:

(i) 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
$\begin{array}{l}
{\left[\begin{array}{ccc}
2 & 2 & -2 \\
4 & 4 & -1 \\
6 & 6 & 2
\end{array}\right]\left[\begin{array}{l}
\mathrm{X} \\
\mathrm{y} \\
\mathrm{Z}
\end{array}\right]=\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]} \\
|\mathrm{A}|=2(14)-2(14)-2(0) \\
|\mathrm{A}|=0
\end{array}$
Therefore, $A$ is singular. Hence, the given system is either inconsistent or it is consistent with infinitely many solution according to as:
$($ Adj $A) \times B \neq 0$ or $($ Adj $A) \times B=0$
Cofactors of A are:
$\begin{array}{l}
C_{11}=(-1)^{1+1} 8+6=14 \\
C_{21}=(-1)^{2+1} 4+12=-16 \\
C_{31}=(-1)^{3+1}-2+8=6 \\
C_{12}=(-1)^{1+2} 8+6=-14 \\
C_{22}=(-1)^{2+1} 4+12=16
\end{array}$
$\begin{array}{l}
C_{32}=(-1)^{3+1}-2+8=-6 \\
C_{13}=(-1)^{1+2} 24-24=0 \\
C_{23}=(-1)^{2+1} 12-12=0 \\
C_{33}=(-1)^{3+1} 8-8=0
\end{array}$
$\begin{array}{l}
\text { Adj } A=\left[\begin{array}{ccc}
14 & -14 & 6 \\
-16 & 16 & -6 \\
0 & 0 & 0
\end{array}\right]^{\mathrm{T}} \\
=\left[\begin{array}{ccc}
14 & -16 & 6 \\
-14 & 16 & -6 \\
0 & 0 & 0
\end{array}\right] \\
\operatorname{Adj} A \times \mathrm{B}=\left[\begin{array}{ccc}
14 & -16 & 6 \\
-14 & 16 & -6 \\
0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]
\end{array}$
Now, $A X=B$ has infinite many solution
Suppose $z=k$
Therefore, $2 x+2 y=1+2 k$
$4 x+4 y=2+k$
We can write this as:
$\left[\begin{array}{ll}
2 & 2 \\
4 & 4
\end{array}\right]\left[\begin{array}{l}
\mathrm{x} \\
\mathrm{y}
\end{array}\right]=\left[\begin{array}{c}
1+2 \mathrm{k} \\
2+\mathrm{k}
\end{array}\right]$
Thus, $|\mathrm{A}|=0 \mathrm{z}=0$
As a result, the given equation doesn’t satisfy.