Answer
See below
Work Step by Step
According to The Adjoint Method for Computing $A^{-1}$, we have the formula:
$$A^{-1}=\frac{1}{\det(A)}adj(A)$$
Since $adj(A)=M^T_C$, we find:
$C _{11}=(-1)^{1+1}.M_{11}=-1\\
C_{21}=(-1)^{2+1}.M_{21}=3\\
C_{31}=(-1)^{3+1}.M_{31}=-2\\
C_{12}=(-1)^{1+2}.M_{12}=2\\
C_{22}=(-1)^{2+2}.M_{22}=-6\\
C_{32}=(-1)^{3+2}.M_{32}=4\\
C_{13}=(-1)^{1+3}.M_{13}=-1\\
C_{23}=(-1)^{2+3}.M_{23}=3\\
C_{33}=(-1)^{3+3}.M_{33}=-2$
Hence, $M_C=\begin{bmatrix}
-1 & 2 & -1 \\ 3 & -6 & 3\\-2 & 4 & -2
\end{bmatrix}\\
\rightarrow adj(A)=\begin{bmatrix}
-1 & 3 & -2\\2 & -6 & 4\\-1 & 3 & -2
\end{bmatrix}$
Consequently, $A.\det(A)=\det(A).I_n\\
\rightarrow \det(A)=\begin{bmatrix}
1 & 2 & 3 \\3 & 4 & 5\\5 & 6 & 7
\end{bmatrix}\begin{bmatrix}
-1 & 3 & -2\\2 & -6 & 4\\-1 & 3 & -2
\end{bmatrix}=\begin{bmatrix}
0 & 0 & 0\\0 & 0 & 0 \\ 0& 0 & 0
\end{bmatrix}=0$
