In a △ABC, it is given that 2tanA=3tanB=4tanC=k
(Let) ⇒tanA=2k,tanB=3k and tanC=4k.
Since in △ABC, tanA+tanB+tanC=tanAtanBtanC ⇒9k=24k3 ⇒k2=83{∵k=0} ⇒9k=24k3 ⇒k2=83{∵k=0}
Now, sec2A+sec2B+sec2C =3+tan2A+tan2B+tan2C =3+k2[4+9+16]=3+83(29)=824+87=8111 =3+tan2A+tan2B+tan2C =3+k2[4+9+16]=3+83(29) =824+87=8111