Interesting sum of $3\times 3$ determinants, using interchanging of rows and columns

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To every $9$ digit number $\overline{abcdefghi}$ we associate the $3\times 3$ matrix: $$ \left[ \begin{array}{ccc} a &b &c \\ d &e &f\\ g &h &i\\ \end{array} \right] $$

Find the sum of determinants of all such matrices.

I am aware of the fact that I need to use the fact that interchanging two rows/columns changes the sign of the determinant, but I need some help on "pairing" the $9$ digit-numbers so that the determinants cancel out somehow.

I have to mention that I found this on a multiple-choice test, and the possible answers are :

a. a number of the form $9$$k$+$1$, with $k$ an integer

b. $999$

c. $-1$

d. $2022$

e. $other \,answer$

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We can see that every arrangement $W$ to put the numbers in the Matrix has 3 other "Symmetric" arrangements :
$X$ , where we rotate it by $90^0$.
$Y$ , where we rotate it by $180^0$.
$Z$ , where we rotate it by $270^0$.
These 4 arrangements (when non-Zero) have Same magnitude but alternating Sign.
Hence , total must be $W+X+Y+Z=W-W+W-W$ which is $0$.

Alternately :
Every arrangement $A$ has a "Partner" arrangement $B$ where Column 1 is exchanged with Column 2.
When these 2 arrangements are not Identical , Determinants must have Same magnitude but alternating Sign.
Hence , total must be $A+B=A-A$ which is $0$.
When these 2 arrangements are Identical , Both Determinants must be 0.
Hence , total must be $A+A=0+0$ which is $0$.

Overall , taking all arrangements , total must be $0+0+0+.....0+0=0$

We can verify this with $2\times2$ Case with Numbers $1,2,3,4$.