The distance between the point (3, 4, 5) and the point where the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ meets the plane x + y + z = 17, is

3

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

⇒ x = λ + 3,  y = 2λ + 4,   z = 2λ + 5

meeting point with plane 

using them in plane is equation (x + y + z = 17)

⇒  λ + 3 + 2λ + 4 + 2λ + 5 = 17

⇒  5λ + 12 = 17

⇒  5λ = 5

⇒  λ = 1

so  x = 1 + 3  ;   y = 2 + 4  ;  2 + 5 = z

x = 4   ;   y = 6   ;    z = 7

given P(3, 4, 5)

so distance = $\sqrt{(4-3)^2+(6-4)^2+(7-5)^2}$

= $\sqrt{1^2+2^2+2^2}$

= $\sqrt{1 + 4 + 4}$

= $\sqrt{9}$

= 3