Permutation in line and circle with conditions.
In a row: There are $10!$ ways to arrange the people. We now count the bad arrangements, in which two A's are next to each other, or two B's, or two C's.
To do the counting, we use Inclusion/Exclusion. How many arrangements have the two A's together? Put them in a bag. We now have $8$ people and $1$ bag. These can be arranged in $9!$ ways. Then when we let the A's out of the bag, they can arrange themselves in $2$ ways, for a total of $2\cdot 9!$.
Do the same for the B's, the C's, and add up. We get $3\cdot 2\cdot 9!$.
However, we have double-counted the bad arrangements in which, for example, the A's and the B's are together. The same bag argument shows there are $2^2\cdot 8!$ such arrangements, for a total of $3\cdot 2^2\cdot 8!$.
Thus our next estimate for the number of bads is $3\cdot 2\cdot 9!-3\cdot 2^2\cdot 8!$.
However, we have subtracted one too many times the arrangements in which the A's are together, and the B's, and the C's. The now familiar technique shows there are $2^3\cdot 7!$ of these. So add back $2^3\cdot 7!$.
We end up with $7!(6\cdot 9\cdot 8-12\cdot 8+8)$, which is $8!\cdot 43$.
This is the number of bad arrangements. The number of good arrangements is $10!-8!\cdot 43$, which is $8!\cdot 47$.
Circular table: We use the convention that two orders are to be considered the same if they differ by a rotation. Assume that one of the "others" is a Canadian. He/she is probably too polite to complain, so one can put the Canadian in the worst chair.
The remaining chairs can now be thought of as a line. We have $9$ people, including $3$ "others" to put in a line, with the condition that the two A's, two B's, and two C's are separated. Same Inclusion/Exclusion technique.
Seating in a row
I shall name the people of $3$ particular nationalities $A_1\;\; A_2\;\; B_1\;\; B_2\;\; C_1\;\; C_2$
First permute $4$ men of "other" nationalities + $A_1,\;B_1,\;C_1\;$ in $7!$ ways.
In the diagram below, $A_1$ is shown in black among the $7$
$\uparrow\huge\circ$$\uparrow\huge\circ$$\uparrow\huge\circ$$\uparrow\huge\circ$$\huge\bullet$$\huge\circ$$\uparrow\huge\circ$$\uparrow$
Wherever $A_1$ may be placed, $A_2$ has only $7-1 = 6 $ places where it can be put.
Similarly, $B_2$ will have $8-1 = 7$ places where it can be put, and $C_3$, $8$ places.
Thus number of arrangements $= 7!\times(6\cdot7\cdot8) = 42*8!$
Addendum
This misses a few possible placements by ignoring ones where a "duplicate" is placed together with the "original." On revisiting the question (after many days !) I have developed a method to work out a way to circumvent this lacuna
Suppose 2 objects are to be kept apart in a permutation, there are $2$ well known methods
the "gap method", where these $2$ objects are placed in the gaps (incl. ends)
the "subtraction method" [ total permutations - those with the $2$ objects together ]
By successively applying these, we can find the number of ways both $A's$ and $B's$ are apart, and then place the $C's$, as seperators, if so needed.
$A's$ apart or together, $B's$ apart $= 6!\times 7\cdot 6 = 6!\times 42$
$(a) A's$ together, $B's$ apart $= (2\times 5!) \times 6\cdot 5 = 6!\times 10$
$(b)$ thus both $A's$ and $B's$ apart = $6!(42-10) = 6!\times 32$
$(c)$ also, both $A's$ and $B's$ together = $6!\times 4$
Now placing the $C's$ appropriately, remembering that the two are distinct, and that there will be two versions of case $(a)$, we get
$6![ (2\times 10) \times (2\cdot 8) + (32\times8\cdot9) + (2\cdot4)] = 8!\times 47$