The distance of the point $$(1, -5, 9)$$ from the plane $$x - y + z = 5$$ measured along the line $$x = y = z$$ is

We have to find the distance from the point $$P(1,-5,9)$$ to the plane $$x-y+z=5$$, but not the perpendicular distance. The question specifies that the measurement must be taken along the line $$x=y=z$$. This line has direction ratios proportional to $$1:1:1$$, so its direction vector can be written as $$\vec d=\langle 1,1,1\rangle$$.

A line that passes through the given point $$P(1,-5,9)$$ and is parallel to $$\vec d$$ can be expressed in parametric form. Let us choose the parameter $$t$$; then every point on this required line is

$$\bigl(x,y,z\bigr)=\bigl(1,-5,9\bigr)+t\langle 1,1,1\rangle=\bigl(1+t,\,-5+t,\,9+t\bigr).$$

The intersection point of this line with the plane must satisfy the plane equation $$x-y+z=5$$. Substituting the coordinates of the general point on the line into the plane equation, we get

$$\bigl(1+t\bigr)\;-\;\bigl(-5+t\bigr)\;+\;\bigl(9+t\bigr)=5.$$

Now we open the brackets and collect like terms one by one:

$$1+t\;+\;5-t\;+\;9+t=5.$$

The $$t$$ terms are $$t - t + t = t$$, and the constant terms are $$1+5+9 = 15$$, so the left-hand side simplifies to

$$15 + t = 5.$$

Solving this simple linear equation for $$t$$ gives

$$t = 5 - 15 = -10.$$

Thus the required point of intersection (let us call it $$Q$$) is obtained by substituting $$t=-10$$ into the parametric coordinates:

$$Q = \bigl(1-10,\,-5-10,\,9-10\bigr)=\bigl(-9,\,-15,\,-1\bigr).$$

The distance between two points lying on a straight line in the direction $$\vec d=\langle 1,1,1\rangle$$ is equal to the absolute value of the parameter difference multiplied by the magnitude of the direction vector. Here the parameter difference is simply $$|t| = |-10| = 10$$, and the magnitude of $$\vec d$$ is

$$|\vec d|=\sqrt{1^2+1^2+1^2}=\sqrt{3}.$$

Therefore, the distance we need is

$$|t|\;|\vec d| = 10\;\sqrt{3}.$$

Hence, the correct answer is Option D.