Let P be the plane which contains the line of intersection of the planes x + y + z - 6 = 0 and 2x + 3y + z + 5 = 0 and it is perpendicular to the xy-plane. Then the distance of the point (0, 0, 256) from P is equal to:

λ(x + y + z - 6) + 2x + 3y + z + 5 = 0
(λ + 2)x + (λ + 3)y + (λ + 1)z + 5 - 6λ = 0
If the plane is perpendicular to the xy-plane, then the coefficient of z must be 0.
λ + 1 = 0 ⇒ λ = -1
Substituting λ = -1 in the equation of the plane:
( -1 + 2)x + (-1 + 3)y + (-1 + 1)z + 5 - 6(-1) = 0
x + 2y + 11 = 0
The distance of the point (0, 0, 256) from the plane x + 2y + 11 = 0 is given by:
Distance = |(1)(0) + (2)(0) + 11| / √(1² + 2² + 0²) = 11/√5