Step 1: Read off the two directions.
The line direction is $\vec d=(1,-1,1)$. The plane $2x+3y+4z=0$ has normal $\vec n=(2,3,4)$.
Step 2: Use the angle formula.
The angle $\theta$ between the line and the normal satisfies
\[ \cos\theta=\frac{|\vec d\cdot\vec n|}{|\vec d|\,|\vec n|}. \]
Step 3: Do the dot product and lengths.
$\vec d\cdot\vec n=2-3+4=3$, $|\vec d|=\sqrt3$, $|\vec n|=\sqrt{29}$.
Step 4: Get $\cos^2\theta$.
\[ \cos^2\theta=\frac{3^2}{3\cdot29}=\frac{9}{87}=\frac{3}{29}. \]
Step 5: Build $\sec^2\theta$ and $\tan^2\theta$.
$\sec^2\theta=\dfrac{1}{\cos^2\theta}=\dfrac{29}{3}$, and $\tan^2\theta=\sec^2\theta-1=\dfrac{29}{3}-1=\dfrac{26}{3}$.
Step 6: Add them.
\[ \tan^2\theta+\sec^2\theta=\frac{26}{3}+\frac{29}{3}=\frac{55}{3}. \]
\[ \boxed{\dfrac{55}{3}} \]