The volume (in cubic units) of the tetrahedro | vedclass

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(C) The equation of the plane is $3x + 4y - 5z = 60$. Dividing by $60$, we get the intercept form: $\frac{x}{20} + \frac{y}{15} - \frac{z}{12} = 1$. T

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$600$

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(C) The equation of the plane is $3x + 4y - 5z = 60$. Dividing by $60$, we get the intercept form: $\frac{x}{20} + \frac{y}{15} - \frac{z}{12} = 1$. This plane intersects the coordinate axes at the points $A(20, 0, 0)$, $B(0, 15, 0)$, and $C(0, 0, -12)$. The tetrahedron is formed by the origin $O(0, 0, 0)$ and the points $A, B, C$. The volume of a tetrahedron with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ and $(x_4, y_4, z_4)$ is given by $V = \frac{1}{6} |\vec{OA} \cdot (\vec{OB} 	imes \vec{OC})|$. Here, $\vec{OA} = 20\hat{i}$, $\vec{OB} = 15\hat{j}$, and $\vec{OC} = -12\hat{k}$. $V = \frac{1}{6} |20\hat{i} \cdot (15\hat{j} 	imes -12\hat{k})| = \frac{1}{6} |20 	imes 15 	imes (-12)| = \frac{1}{6} |-3600| = 600$ cubic units.

The volume (in cubic units) of the tetrahedron bounded by the plane $3x + 4y - 5z = 60$ and the three coordinate planes is

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