If the plane 2x + 3y + z = 6 cuts the coordin | vedclass

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(B) Step $1$: Convert the plane equation to intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Step $2$: Given $2x + 3y + z = 6$, divide by

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$6$

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(B) Step $1$: Convert the plane equation to intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$.Step $2$: Given $2x + 3y + z = 6$, divide by $6$ to get $\frac{2x}{6} + \frac{3y}{6} + \frac{z}{6} = 1$, which simplifies to $\frac{x}{3} + \frac{y}{2} + \frac{z}{6} = 1$.Step $3$: The intercepts are $a = 3$, $b = 2$, and $c = 6$. Thus, the coordinates of $A, B, C$ are $(3, 0, 0)$, $(0, 2, 0)$, and $(0, 0, 6)$.Step $4$: The volume of a tetrahedron with vertices at the origin and intercepts $a, b, c$ is given by $V = \frac{1}{6} |abc|$.Step $5$: $V = \frac{1}{6} 	imes 3 	imes 2 	imes 6 = 6 	ext{ cubic units}$.

If the plane $2x + 3y + z = 6$ cuts the coordinate axes at $A$, $B$, and $C$, then the volume of the tetrahedron $OABC$ (where $O$ is the origin) in cubic units is:

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