The distance of the point $\left(1, 0, 2\right)$ from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane $x-y+z=16$, is 

Let $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=t$  

General points is $(2+3t,- 1+4t,2+12t)$

It lies on the plane $x-y+z=16.$

$\Rightarrow 2+3t+1-4t+2+12t=16$

$\Rightarrow t=1$

Hence, the point of intersection will be

$(2+3(1),-1+4(1),2+12(1\left)\right)=\left(5,3,14\right)$

Distance from $(1,0,2)$ $=\sqrt{{\left(5-1\right)}^2+{\left(3-0\right)}^2+{\left(14-2\right)}^2}$

$=\sqrt{4^2+3^2+12^2}$

$=13$