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

The correct answer is option (3) → 3

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

$⇒x=λ+3,y=2λ+4,z=2λ+5$

placing in eq. of plane

$λ+3+2λ+4+2λ+5=17$

$5λ=5⇒λ=1$

$x=4,y=6,z=7$ → points of intersection

given point (3, 4, 5)

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

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

$=3$