Given equation of line is
$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda$ [say] ...(i)
and equation of plane is
$x-y+z=16$
Any point on the line (i) is
$(3 \lambda+2,4 \lambda-1,12 \lambda+2)$
Let this point of intersection of the line and plane.
$\therefore(3 \lambda+2)-(4 \lambda-1)+(12 \lambda+2)=16$
$11 \lambda+5=16$
$11 \lambda=11 \Rightarrow \lambda=1$
So, the point of intersection is $(5,3,14)$.
Now, distance between the points $(1,0,2)$ and $(5,3,14)$
$=\sqrt{(5-1)^{2}+(3-0)^{2}+(14-2)^{2}} $
$=\sqrt{16+9+144}=\sqrt{169}=13$