Q.

If the points P and Q are respectively the circumcenter and the orthocentre of a ABC, then PA+PB+PC is             [2023]

1 2PQ  
2 QP  
3 2QP  
4 PQ  

Ans.

(4)

PA+PB+PC=a+b+c

=3(a+b+c)3

=3PG=PQ