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

Equation of line parallel to $x=y=z$ through

$\left(1,-5, 9\right)$ is $\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda$

If $\mathrm{P }\left(\lambda +1, \lambda -5, \lambda +9\right)$ be point of intersection of line and plane.

$\Rightarrow$ $\lambda +1-\lambda +5+\lambda +9=5$

$\Rightarrow$ $\lambda = -10$

$\Rightarrow$ Coordinate points are $\left(-9, -15, -1\right)$

$\Rightarrow$ Required distance $=\left|\lambda \right|\sqrt{1^2+1^2+1^2}=10\sqrt{3}$