The distance of the point (1, 1, 9) from the point of intersection of the line  $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z+5}{2}$ and the plane $x+y+z=17$ is:       

The given line is $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda$

The general point on the line is $\Rightarrow x=\lambda +3,y=2\lambda +4,z=2\lambda +5$

This point also lies on the plane 

hence, $\Rightarrow \lambda +3+2\lambda +4+2\lambda +5=17\Rightarrow \lambda =1$

The point of intersection is $Q\left(4,6,7\right)$

Hence, the distance $PQ=\sqrt{9+25+4}=\sqrt{38}$