Let (a-d), a, and (a+d) be the three numbers in Arithmetic Series.

Given,

1. The sum of the three numbers is 27. $\rm (a - d) + a + (a + d) = 27$ $\rm \Rightarrow 3a = 27 \Rightarrow a = 9$.
2. The product of the three numbers is 585. $\rm (a - d) \cdot a \cdot (a + d) = 585$.

On solving the second condition, we get,

$\rm or, a \cdot (a - d) (a + d) = 585$

By using the formula, $\rm (a - b)(a + b) = a^{2} - b^{2}$, we get,

$\rm or, a \cdot (a^{2} - d^{2} ) = 585$

By substituting the value of a obtained above as $\rm a = 9$ in the equation, we get,

$\rm or,  9 \cdot ( 9^{2} - d^{2} ) = 585$

$\rm or, (9^{2} - d^{2}) = \frac{585}{9}$

$\rm or, 9^{2} - d^{2} = 65$

$\rm or, 81 - d^{2} = 65$

$\rm or, 81 + d^{2} - d^{2} = 65 + d^{2}$

$\rm or, 81 = 65 + d^{2}$

$\rm or, 81 - 65 = 65 - 65 + d^{2}$

$\rm or, 16 = d^{2}$

$\rm or, d = \pm \sqrt{16}$

$\rm \therefore d = \pm 4$

When we take $\rm d = 4$, we get,

$\rm (a - d) = (9 - 4) = 5; a = 9; (a + d) = (9 + 4) = 13$

When we take $\rm d = -4$, we get,

$\rm (a - d) = (9 - (-4)) = 13; a = 9; (a + d) = (9 + (-4)) = 5$

Hence, the required numbers are {5,9,13}.