Let, the point of intersection of the line

and the plane x + y + z=17 be (x₀, y₀, z₀).

As (x₀, y₀, z₀) is the point of intersection of the line and

the plane, so it must satisfy both of the equation of

line and the equation of plane.

Substituting, (x₀, y₀, z₀) in place of (x, y, z) in both the equations, we get,

i.e. x₀=k + 3,

y₀=2k + 4 and

z₀=2k + 5

Putting this values in the equation of plane we get,

x₀ + y₀ + z₀=17

(k + 3) + (2k + 4) + (2k + 5)=17

5k + 12=17

5k=5

k=1

∴ x₀=k + 3

=1 + 3

=4

y₀=2k + 4

=(2×1) + 4

=6

z₀=2k + 5

=(2×1) + 5

=7

Hence, the point of intersection is, (4, 6, 7).

Now, the distance between the point (3, 4, 5) and (4, 6, 7) is,

Hence, the distance between the point (3, 4, 5) the point where the line meets the plane x + y + z = 17, is 3 units.