Let $P$ be foot of perpendicular of point $Q\left(1,2,4\right)$

Let $\frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}=\lambda$

So $x=4\lambda -2, y=2\lambda +1,z=3\lambda -1$

Then coordinates of point $P$ will be $\left(4\lambda -2, 2\lambda +1,3\lambda -1\right)$

Now direction ratios of $QP$ 

and D.R's of line will be 

Now $PQ$ and line are perpendicular

so, $4\left(4\lambda -3\right)+2\left(2\lambda -1\right)+3\left(3\lambda -5\right)=0$

$\Rightarrow \lambda =1$

Putting the value of $\lambda$ in $P$ we get $P\equiv \left(2,3,2\right)$

Now distance of point $P\left(2,3,2\right)$ from plane$3x+4y+12z+23=0$ will be  $=\left|\frac{3\times 2+4\times 3+12\times 2+23}{\sqrt{9+16+144}}\right|=\frac{65}{13}$