Sum of a three-digit number and its mirror image is a multiple of $111$. What is this sum (provided 6 is one of the digits of this three-digit number)?

(i) If $(2{)}_{\mathrm{N}}\times (4{)}_{\mathrm{N}}=(8{)}_{\mathrm{N}}$, where $(\mathrm{X}{)}_{\mathrm{N}}$ is a number written in a system of writing having N digits, then N can have infinite values.

(ii) If $(4{)}_{\mathrm{N}}\times (5{)}_{\mathrm{N}}=(24{)}_{\mathrm{N}}$, where $(\mathrm{X}{)}_{\mathrm{N}}$ is a number written in a system of writing having N digits, then N will have more than $1$, but finite values.

(iii) If $(5{)}_{\mathrm{N}}\times (6{)}_{\mathrm{N}}=(3\mathrm{A}{)}_{\mathrm{N}}$, where $(\mathrm{X}{)}_{\mathrm{N}}$ is a number written in a system of writing having N digits and A is the units digit of this number $\left(3\mathrm{A}\right)$ written in a system of writing the numbers having N digits, then A can have four values.

What will be the tens place digit in the following expression:

$1!+2!+3!+\dots +70!?$