Official solution
Here, \[P\left( {\overline A \cap \overline B \left| C \right.} \right) = {{P\left( {\overline A \cap \overline B \cap C} \right)} \over {P\left( C \right)}}\]
= \[{{P\left[ {\left( {\overline {A \cup B} } \right) \cap C} \right]} \over {P\left( C \right)}}\]
= \[{{P\left[ {C - \left( {A \cup B} \right)} \right]} \over {P\left( C \right)}}\]
= \[{{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right) + P\left( {A \cap B \cap C} \right)} \over {P\left( C \right)}}\]
= \[{{P\left( C \right) - P\left( {A \cap C} \right) - P\left( {B \cap C} \right)} \over {P\left( C \right)}}\] (\[ \because \] \left. {P\left( {A \cap B \cap C} \right) = 0} \right) (
= \]{{P\left( C \right) - P\left( A \right).P(C) - P\left( B \right).P(C)} \over {P\left( C \right)}}\[
[\] \because \[ A, B and C are independent events]
= 1 - P(A) - P(B)
= \]P\left( {\overline A } \right) ( - P(B) or \]P\left( {\overline B } \right)