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 :
Answer & explanation
Correct answer: option 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 = $\sqrt{(4-3)^2+(6-4)^2+(7-5)^2}$
$=\sqrt{1^2+2^2+2^2}$
$=3$