If the line $3 x+4 y-24=0$ intersects $X$ and $Y$ axes in points $A$ and $B$ respectively then incenter of…

Incenter of triangle $OAB$ with vertices at intercepts of line $3x+4y-24=0$

The line $3x+4y-24=0$ intersects the axes at $A(8,0)$ (x-intercept) and $B(0,6)$ (y-intercept), with $O=(0,0)$.

Triangle $OAB$ is right-angled at $O$ with side lengths: $OA=8$, $OB=6$, and hypotenuse $AB=10$.

Using the incenter coordinate formula with $a=10$ opposite $O$, $b=6$ opposite $A$, and $c=8$ opposite $B$:

$x = \frac{10\cdot0 + 6\cdot8 + 8\cdot0}{10+6+8} = \frac{48}{24} = 2$

$y = \frac{10\cdot0 + 6\cdot0 + 8\cdot6}{10+6+8} = \frac{48}{24} = 2$

Alternatively, for a right triangle with legs $8$ and $6$, the inradius is $r = \frac{8+6-10}{2} = 2$, confirming the incenter lies at $(r,r)=(2,2)$.

Final answer: $\boxed{(2,2)}$