Given, P(1,1,9). Equation of plane x+y+z=17 Equation of line ⇒
x−3
1
=
y−4
2
=
z−5
2
⇒
x−3
1
=
y−4
2
=
z−5
2
=λ (let) ⇒x=λ+3;y=2λ+4;z=2λ+5 ∴ The point we have is (λ+3,2λ+4,2λ+5). ∵ This point lies on the plane x+y+z=17. ∴λ+3+2λ+4+2λ+5=17 ⇒λ=1 ∴ The coordinate of point is (4,6,7) ∴ Required distance between (1,1,9) and (4,6,7) is =√(4−1)2+(6−1)2+(7−9)2 =√9+25+4=√38
