Let’s consider the three terms of the A.P. to be $a–d$, a, $a+d$

Therefore, the sum of three terms $=21$

⇒ $a–d+a+a+d=21$

⇒ $3a=21$

⇒ $a=7$ It is already given the  product of the first and 3rd = 2nd term + 6

Therefore,

⇒ $(a–d)(a+d)=a+6$

${{a}^{2}}-{{d}^{2}}=a+6$

⇒ ${{\left( 7 \right)}^{2}}-{{d}^{2}}=7+6$

⇒ $49-{{d}^{2}}=13$

⇒ ${{d}^{2}}=49-13=36$

⇒ ${{d}^{2}}={{\left( 6 \right)}^{2}}$

⇒ $d=6$

Hence, the terms are $7–6$, $7$, $7+6⇒1$, $7$, $13$