The sum of three numbers in AP is 3 their product is −35.

Let the required numbers be (a - d), a and (a + d). 
Then (a - d) + a + (a + d) = 3
⇒ 3a = 3
⇒ a = 1​
Also, (a - d).a.(a + d)​ = -35

$\Rightarrow a\left(a^{2}-d^{2}\right)=-35$

$\Rightarrow 1 \cdot\left(1-d^{2}\right)=-35$

$\Rightarrow d^{2}=36$

$\Rightarrow d=\pm 6$

Thus, $a=1$ and $d=\pm 6$

 Hence, the required numbers are ( -5, 1 and  7) or ( 7, 1 and -5).