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Here is a step by step computation:
!a&!b|!c = !a&!b&c|!c = !a&!b&c⊕!c = (a⊕true)&(b⊕true)&c ⊕ (c⊕true) = (a⊕true)&(b⊕true)&c ⊕ c ⊕ true = (a&b ⊕ a ⊕ b ⊕ true)&c ⊕ c ⊕ true = (a&b&c ⊕ a&c ⊕ b&c ⊕ c) ⊕ c ⊕ true = a&b&c ⊕ a&c ⊕ b&c ⊕ c ⊕ c ⊕ true = a&b&c ⊕ a&c ⊕ b&c ⊕ true