The distances of the point $(1,-5,9)$ from the plane $x+y+z=5$ measured along a straight line $x=y=z$ is $2\sqrt{3}k$, then the value of $k$ is

Given equation of plane is $x+y+z=5$.

The distance measured along the line $x=y=z$.

Direction ratios of the given line is $(1,1,1)$.

So, the equation of line $PQ$ is

$\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda$

$\Rightarrow x=\lambda +1,y=\lambda -5,z=\lambda +9$

lies on the plane $x+y+z=5$

$\therefore (\lambda +1)-(-\lambda +5)+(\lambda +9)=5$

$\Rightarrow \lambda =-10$

The coordinate of $Q$ is $(-9,-15,-1)$ and the coordinate of $P$ is $(1,-5,9)$

$PQ=\sqrt{(10{)}^2+(10{)}^2+(10{)}^2}=10\sqrt{3}$

$\therefore 2\sqrt{3}k=10\sqrt{3}\Rightarrow k=5$