The steps taken are correct, but in the distributive law step there is a missing part:

(a&b&c) ⊕ (a&b&c) ⊕ (a&c) ⊕ (a&b&c) ⊕ (a&b) ⊕ (a&c) ⊕ a ⊕ (a&b&c) ⊕ (a&b) ⊕ (a&c) ⊕ (b&c) ⊕ a ⊕ b ⊕ c **⊕ 1.**

From the last bracket ((a ⊕ 1) & (b ⊕ 1) & (c ⊕ 1)) you get (a&b&c) ⊕ (a&b) ⊕ (a&c) ⊕ (b&c) ⊕ a ⊕ b ⊕ c ⊕ 1

On the next step of deleting doubles entries, I think there is misunderstanding. If you have double entries that term will be deleted based on:

a ⊕ a = 0

0 ⊕ a = a

Based on the syntactic sugar above and if we rearrange the term a bit, we will have:

(a&b&c) ⊕ (a&b&c) ⊕ (a&b&c) ⊕ (a&b&c) ⊕ (a&c) ⊕ (a&c) ⊕ (a&c) ⊕ (a&b) ⊕ (a&b) ⊕ (b&c) ⊕ a ⊕ a ⊕ b ⊕ c ⊕ 1

= 0 ⊕ (a&c) ⊕ 0 ⊕ (b&c) ⊕ 0 ⊕ b ⊕ c ⊕ 1 = (a&c) ⊕ (b&c) ⊕ b ⊕ c ⊕ 1.