How to evaluate the volume of tetrahedron bounded between coordinate planes and tangent plane? Find the volume of the tetrahedron in $\mathbb{R}^3$ bounded by the coordinate planes $x =0, y=0, z=0$, and the
tangent plane at the point $(4,5,5)$ to the sphere $(x -3)^2 +(y -3)^2 +(z -3)^2 = 9$.
My attempt: I started with determining the equation of tangent plane which comes out to be $x+2y+2z=24$. This is because direction ratios of normal to sphere at $(4, 5, 5)$ are $2, 4, 4$. So, then equation of tangent plant is given by $2(x-4)+4(y-5)+4(z-5)=0$ which means $x+2y+2z=24$.
The required volume is $$\int _{x=0}^4\int _{y=0}^{12-\frac{x}{2}}\int _{z=0}^{12-y-\frac{x}{2}}\:\:dz\:dy\:dx$$ but this is not giving me the required answer which is $576$. Please help.

Your work seems fine, there is only an issue for the $x$ upper limit
$$V=\int _{x=0}^{24}\int _{y=0}^{12-\frac{x}{2}}\int _{z=0}^{12-y-\frac{x}{2}}\:\:dz\:dy\:dx$$
As noticed in the comments, to check the result we can use that
$$V=\frac13 Sh$$
which in this case leads to
$$V=\frac13 144 \cdot 12=576$$
