The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6,

Let the three terms of the A.P. be $a-d, a, a+d$.

Then, we have:

$a-d+a+a+d=21$

$\Rightarrow 3 a=21$

$\Rightarrow a=7 \ldots(i)$

Also, $(a-d)(a+d)-a=6$

$\Rightarrow a^{2}-d^{2}-a=6$

$\Rightarrow 49-d^{2}-7=6$

$\Rightarrow 36=d^{2}$

$\Rightarrow \pm 6=d$

When $\mathrm{d}=6, \mathrm{a}=7$, we get:

$1,7,13$

When $\mathrm{d}=-6$, a $=7$, we get:

$13,7,1$