In a $\triangle A B C$, it is given that
$\frac{\tan A}{2}=\frac{\tan B}{3}=\frac{\tan C}{4}=k$
(Let) $\Rightarrow \tan A=2 k, \tan B=3 k$ and $\tan C=4 k .$
Since in $\triangle A B C$,
$\tan A+\tan B+\tan C=\tan A \tan B \tan C$
$\Rightarrow 9 k=24 k^{3}$
$\Rightarrow k^{2}=\frac{3}{8}\{\because k \neq 0\}$
$\Rightarrow 9 k=24 k^{3}$
$\Rightarrow k^{2}=\frac{3}{8}\{\because k \neq 0\}$
Now, $\sec ^{2} A+\sec ^{2} B+\sec ^{2} C$
$=3+\tan ^{2} A+\tan ^{2} B+\tan ^{2} C$
$=3+k^{2}[4+9+16]=3+\frac{3}{8}(29)=\frac{24+87}{8}=\frac{111}{8}$
$=3+\tan ^{2} A+\tan ^{2} B+\tan ^{2} C$
$=3+k^{2}[4+9+16]=3+\frac{3}{8}(29)$
$=\frac{24+87}{8}=\frac{111}{8}$