The volume of the tetrahedron included between the plane $3 x+4 y-5 z-60=0$ and the coordinate planes is

The given equation of the plane is $3 x+4 y-5 z-60=0$. It can be written in the form $\frac{x}{20}+\frac{y}{15}+\frac{z}{-12}=1$ which meets the coordinate axes at the pointsA $(20,0,0), \mathrm{B}(0,15,0)$ and $\mathrm{C}(0,0,-12)$. The coordinates of the origin are $\mathrm{O}(0,0,0)$. Therefore, volume of the tetrahedron OABC is $=$ $=\frac{1}{6}\left\|\begin{array}{ccc} 20 & 0 & 0 \\ 0 & 15 & 0 \\ 0 & 0 & -12 \end{array}\right\|=\frac{1}{6}|20 \times 15 \times(-12)|=600$