Official solution
Given, P(1, 1, 9).
Equation of plane x + y + z = 17
Equation of line \[\Rightarrow\] \[{{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2}\]
\[\Rightarrow\] \[{{x - 3} \over 1} = {{y - 4} \over 2} = {{z - 5} \over 2} = \lambda \] (let)
\[\Rightarrow\] x = \[\lambda\] + 3; y = 2\[\lambda\] + 4; z = 2\[\lambda\] + 5
\[\therefore\] The point we have is (\[\lambda\] + 3, 2\[\lambda\] + 4, 2\[\lambda\] + 5).
\[\because\] This point lies on the plane x + y + z = 17.
\[\therefore\] \[\lambda\] + 3 + 2\[\lambda\] + 4 + 2\[\lambda\] + 5 = 17
\[\Rightarrow\] \[\lambda\] = 1
\[\therefore\] The coordinate of point is (4, 6, 7)
\[\therefore\] Required distance between (1, 1, 9) and (4, 6, 7) is
\[ = \sqrt {{{(4 - 1)}^2} + {{(6 - 1)}^2} + {{(7 - 9)}^2}} \]
\[ = \sqrt {9 + 25 + 4} = \sqrt {38} \]