The distance of the point (1,1,9) from the point

Let $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=t$

$\Rightarrow \quad x=3+t, y=2 t+4, z=2 t+5$

for point of intersection with $x+y+z=17$

$3+t+2 t+4+2 t+5=17$

$\Rightarrow 5 \mathrm{t}=5 \Rightarrow \mathrm{t}=1$

$\Rightarrow$ point of intersection is $(4,6,7)$

distance between $(1,1,9)$ and $(4,6,7)$

is $\sqrt{9+25+4}=\sqrt{38}$