If $n\left(A\right)=12,n\left(B\right)=16,n\left(C\right)=21,n(A\cap B)=5,n(A\cap C)=8,n(B\cap C)=12,n(A\cap B\cap C)=3$, then $n(A\cup B\cup C)=$

Given $n\left(A\right)=12,n\left(B\right)=16,n\left(C\right)=21,n(A\cap B)=5,n(A\cap C)=8,n(B\cap C)=12,n(A\cap B\cap C)=3$

We know that, $n(A\cup B\cup C)=n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)$

Substitute the appropriate values in the above formula 

we get 

$\begin{array}{rcl}n(A\cup B\cup C)& =& 12+16+21-5-12-8+3\\ & =& \boxed{27}\end{array}$