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The volume (in cubic units) of the tetrahedran bounded by the plane $3 x+4 y-5 z=60$ and the three coordinate planes is
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600
Vertices of the tetrahedron are $(0,0,0)$,
$(20,0,0),(0,15,0)$ and $(0,0,12)$.
So, its volume is $\frac{1}{6}\left|\begin{array}{cccc}0 & 0 & 0 & 1 \\ 20 & 0 & 0 & 1 \\ 0 & 15 & 0 & 1 \\ 0 & 0 & 12 & 1\end{array}\right|=600$ cubic units
$(20,0,0),(0,15,0)$ and $(0,0,12)$.
So, its volume is $\frac{1}{6}\left|\begin{array}{cccc}0 & 0 & 0 & 1 \\ 20 & 0 & 0 & 1 \\ 0 & 15 & 0 & 1 \\ 0 & 0 & 12 & 1\end{array}\right|=600$ cubic units
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