The distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along the line x = y = z is
10√3
Equation of line passing through the point (1, -5, 9) and parallel to x = y = z is
x−11=y+51=z−91=λ
Thus, any point on this line is of the form
λ+1,λ−5,λ+9.
Now, if P(λ+1,λ−5,λ+9) is the point of intersection of line and plane, then
(λ+1)−(λ−5)+λ+9=5
⇒λ+15=5⇒λ=−10
∴ Coordintes of point P are (-9, -15, -1).
Hence, required distance
=√(1+9)2+(−5+15)2+(9+1)2
=√102+102+102=10√3