The volume of the tetrahedron formed by the p | vedclass

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(C) The given planes are:$y + z = 0$ ... $(1)$$z + x = 0$ ... $(2)$$x + y = 0$ ... $(3)$$x + y + z = 2$ ... $(4)$To find the vertices of the tetrahedr

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$\frac{16}{3}$

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(C) The given planes are:$y + z = 0$ ... $(1)$$z + x = 0$ ... $(2)$$x + y = 0$ ... $(3)$$x + y + z = 2$ ... $(4)$To find the vertices of the tetrahedron,we solve the systems of equations formed by taking three planes at a time:- Intersection of $(1), (2), (3)$: $x+y=0, y+z=0, z+x=0 \implies x=y=z=0$. Vertex $V_1 = (0, 0, 0)$.- Intersection of $(1), (2), (4)$: $y+z=0, z+x=0, x+y+z=2$. Substituting $y=-z$ and $x=-z$ into $(4)$,we get $-z-z+z=2 \implies z=-2$. Thus,$x=2, y=2$. Vertex $V_2 = (2, 2, -2)$.- Intersection of $(1), (3), (4)$: $y+z=0, x+y=0, x+y+z=2$. Substituting $z=-y$ and $x=-y$ into $(4)$,we get $-y-y+y=2 \implies y=-2$. Thus,$x=2, z=2$. Vertex $V_3 = (2, -2, 2)$.- Intersection of $(2), (3), (4)$: $z+x=0, x+y=0, x+y+z=2$. Substituting $z=-x$ and $y=-x$ into $(4)$,we get $x-x-x=2 \implies x=-2$. Thus,$y=2, z=2$. Vertex $V_4 = (-2, 2, 2)$.The volume of the tetrahedron with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3), (x_4, y_4, z_4)$ is given by $\frac{1}{6} |det(V_2-V_1, V_3-V_1, V_4-V_1)|$.Volume $= \frac{1}{6} \left| \begin{matrix} 2 & 2 & -2 \ 2 & -2 & 2 \ -2 & 2 & 2 \end{matrix} ight| = \frac{1}{6} |2(-4-4) - 2(4+4) - 2(4-4)| = \frac{1}{6} |-16 - 16 - 0| = \frac{32}{6} = \frac{16}{3}$.

The volume of the tetrahedron formed by the planes whose equations are $y + z = 0, z + x = 0, x + y = 0$ and $x + y + z = 2$ is

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