Let $A(0,0,0), B(3,0,0), C(0,4,0)$


Now,
$$
\begin{aligned}
& a=\sqrt{(0-3)^2+(4-0)^2+(0-0)^2}=B C \\
& a=\sqrt{25}=5
\end{aligned}
$$
and $b=A C=\sqrt{0+4^2+0}=4$
and $c=A B=\sqrt{3^2+0+0}=3$
Now, coordinate of incentre $(x, y, z)$ is given as
$$
x=\frac{a x_1+b x_2+c x_3}{a+b+c}
$$

$$
\begin{aligned}
& =\frac{15(0)+(4)(3)+(3)(0)}{5+4+3}=1 \\
y & =\frac{a y_1+b y_2+c y_3}{a+b+c}=\frac{(5)(0)+(4)(0)+(3)(4)}{5+4+3}=1 \\
z & =\frac{a z_1+b z_2+c z_3}{a+b+c} \\
& =\frac{(5)(0)+(4)(0)+(3)(0)}{5+4+3}=0 \\
\therefore(x, y, z) & =(1,1,0)
\end{aligned}
$$