$(iii)$ is clealy $\text{incorrect}$. Example :$\begin{bmatrix} 1 &0 &0 \\ 0&0 &1 \\ 0&1 &0 \end{bmatrix}$
Determinant of this matrix is $-1$
Let the matrix $A=$ $\begin{bmatrix} x_{11} &x_{12} &x_{13} &\ldots \\ x_{21} &x_{22} &x_{23} &\ldots \\ x_{31} &x_{32} &x_{33} &\ldots \\ \ldots &\ldots &\ldots &\ldots \\ \end{bmatrix}$ of order $n*n$
$\sum_{j=1}^{n}x_{ij}=1\,\,\forall\,\,i=1,\ldots,n$
We know, $Av = \lambda v$, where $\lambda$ are the eigenvalues and $v$ is the eigenvector
As, $Av=v$
$\begin{bmatrix} x_{11} &x_{12} &x_{13} &\ldots \\ x_{21} &x_{22} &x_{23} &\ldots \\ x_{31} &x_{32} &x_{33} &\ldots \\ \ldots &\ldots &\ldots &\ldots \\ \end{bmatrix} \begin{bmatrix} 1\\ 1\\ 1\\ ...\\ \end{bmatrix} = \begin{bmatrix} x_{11}+x_{12}+x_{13}+\ldots\\ x_{21}+x_{22}+x_{23}+\ldots\\ x_{31}+x_{32}+x_{33}+\ldots\\ x_{41}+x_{42}+x_{43}+\ldots\\ \end{bmatrix}=\begin{bmatrix} 1\\ 1\\ 1\\ \ldots \\ \end{bmatrix}=v$
$v$ is an eigen vector and eigen value is $1$
So, $(i)$ is $correct$
$A\,-\,\lambda I = \begin{bmatrix} x_{11}-1 &x_{12} &x_{13} &\ldots \\ x_{21} &x_{22}-1 &x_{23} &\ldots \\ x_{31} &x_{32} &x_{33}-1 &\ldots \\ \ldots &\ldots &\ldots &\ldots \\ \end{bmatrix}$
Now, $\sum_{j=1}^{n}x_{ij}=0\,\,\forall\,\,i=1,\ldots,n$
Applying Matrix transformation,
$C_{1} \leftarrow\sum_{j=1}^{n}C_{j}$
The matrix now becomes = $ \begin{bmatrix} 0 &x_{12} &x_{13} &\ldots \\ 0 &x_{22}-1 &x_{23} &\ldots \\ 0 &x_{32} &x_{33}-1 &\ldots \\ \ldots &\ldots &\ldots &\ldots \\ \end{bmatrix}$
As $C_{1}$ is completely 0 , so $|A-\lambda I|=0$
$(ii)$ is $correct$
So, option (C).