The incentre of the tetrahedron formed by the planes $x=0,y=0,z=0$  and $x+y+z=2024$  is.

It is evident that the planes $x=0,y=0,z=0$ meet in $\left(0,0,0\right)$ .Hence the incentre lies on the perpendicular from $\left(0,0,0\right)$ to the plane $x+y+z=a$   and divides it in the ratio 3:1 ( 3 from the vertex $\left(0,0,0\right)$ and 1from the plane $x+y+z=a$ ).
The equation of the perpendicular from $\left(0,0,0\right)$  to the plane $x+y+z=a$  is $\frac{x}{1}=\frac{y}{1}=\frac{z}{1}=r\left(say\right)$                                             
Any point this perpendicular is  $\left(r,r,r\right)$.If it lies on the plane $x+y+z=a$ ,then we have   $r+r+r=a$
Or        $r=\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
$\therefore$ The perpendicular from  $\left(0,0,0\right)$ meets the plane $x+y+z=a$   in $\left(r,r,r\right)$  i.e., $\left(\frac{a}{3},\frac{a}{3},\frac{a}{3}\right)$  .Also the incentre divides the join of $\left(0,0,0\right)$  and $\left(\frac{a}{3},\frac{a}{3},\frac{a}{3}\right)$  in the ratio 3:1 , therefore if $\left(x_1,y_1,z_1\right)$   be the required incentre, 
We have     $x_1=\frac{3.\frac{a}{3}+1.0}{3+1}=\frac{a}{4}$
Similarly  $y_1=\frac{a}{4}=z$
$\therefore$The required incentre is $\left(\frac{a}{4},\frac{a}{4},\frac{a}{4}\right)$ . Since  $a=2024$  then incentre is  $\left(506,506,506\right)$