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.

The plane intercept the axes in the points
\begin{align}
A&=(24,0,0), \\
B&=(0,12,0), \\
C&=(0,0,12)
\end{align}
so the volume is
$$
V=\frac{1}{6}\cdot24\cdot12\cdot12=576
$$
