Thank you for reporting, we will resolve it shortly
Q.
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 $P$ from the plane $3 x+4 y+12 z+23=0$
$\frac{ x +2}{4}=\frac{ y -1}{2}=\frac{ z +1}{3}=\lambda$
$( x , y , z )=(4 \lambda-2,2 \lambda+1,3 \lambda-1)$
$\overrightarrow{ AP }=(4 \lambda-3) \hat{ i }+(2 \lambda-1) \hat{ j }+(3 \lambda-5) \hat{ k }$
$\vec{ b }=4 \hat{ i }+2 \hat{ j }+3 \hat{ k }$
$\overrightarrow{ AP } \cdot \vec{ b }=0$
$4(4 \lambda-3)+2(2 \lambda-1)+3(3 \lambda-5)=0$
$29 \lambda=12+2+15=29$
$\lambda=1$
$P =(2,3,2)$
$3 x +4 y +12 z +23=0$
$d =\left|\frac{6+12+24+23}{\sqrt{9+16+144}}\right|$
$d =\left|\frac{65}{13}\right|=5$
