The sum of three numbers in $A.P.$ is $21$. The product of the first and the third number exceeds the second number by $6$. Find those numbers.

(A) Let the three numbers in $A.P.$ be $(a - d)$,$a$,and $(a + d)$.
According to the problem,the sum of these numbers is $21$:
$(a - d) + a + (a + d) = 21$
$3a = 21$
$a = 7$
Now,the product of the first and the third number exceeds the second number by $6$:
$(a - d)(a + d) = a + 6$
$a^2 - d^2 = a + 6$
Substitute $a = 7$ into the equation:
$7^2 - d^2 = 7 + 6$
$49 - d^2 = 13$
$d^2 = 36$
$d = \pm 6$
Case $1$: If $d = 6$,the numbers are $(7 - 6), 7, (7 + 6)$,which are $1, 7, 13$.
Case $2$: If $d = -6$,the numbers are $(7 - (-6)), 7, (7 + (-6))$,which are $13, 7, 1$.
Thus,the numbers are $1, 7, 13$ or $13, 7, 1$.