Let the foot of the perpendicular from the point $(1,2,4)$ on the line $\frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}$ be P. Then the distance of $\mathrm{P}$ from the plane $3x+4y+12z+23=0$

 $L:\frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}=t$
Let $P=(4t-2,2t+1,3t-1)$
$\because P$ is the foot of perpendicular of $A(1,2,4)$

drs of PA(4t-2-1,2t+1-2,3t-1-4)=$(4t-3, 2t-1,3t-5)=0$
$\therefore 4(4t-3)+2(2t-1)+3(3t-5)=0$
$\Rightarrow 29t=29\Rightarrow t=1$
$\therefore P=(2,3,2)$
Now, distance of $P$ from the plane
$3x+4y+12z+23=0$, is
$\left|\frac{6+12+24+23}{\sqrt{9+16+144}}\right|=\frac{65}{13}=5$