Given,

$a,b,c$ are three particular speakers among the $10$ speakers of a meeting.

So number of ways of arranging $10$ speaker will be $10!$

Now assuming all three $a, b \& c$ are sitting together, so number of ways of arranging them will be $3\times 8!$

Now the number of ways of arranging all the $10$ speakers on the dias in a row so that all the three speakers $a,b,c$ do not sit together is Total ways $-$ all sitting together $=10!-3\times 8!=8!\left(90-6\right)=84\times 8!$