Consider the tetrahedron with the vertices A( | vedclass

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(C) The vertices of the tetrahedron are $A(3,2,4)$,$B(x_1, y_1, 0)$,$C(x_2, y_2, 0)$,and $D(x_3, y_3, 0)$.Triangle $BCD$ is formed by the intersection

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$\left(3, \frac{7}{4}, 1\right)$

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(C) The vertices of the tetrahedron are $A(3,2,4)$,$B(x_1, y_1, 0)$,$C(x_2, y_2, 0)$,and $D(x_3, y_3, 0)$.Triangle $BCD$ is formed by the intersection of lines $y=x$,$x+y=6$,and $y=1$.Solving for the intersection points:$1$) $y=x$ and $y=1 \Rightarrow x=1$. Vertex $B = (1,1,0)$.$2$) $x+y=6$ and $y=1 \Rightarrow x=5$. Vertex $D = (5,1,0)$.$3$) $y=x$ and $x+y=6 \Rightarrow 2x=6 \Rightarrow x=3, y=3$. Vertex $C = (3,3,0)$.The centroid $G$ of a 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 $\left(\frac{\sum x_i}{4}, \frac{\sum y_i}{4}, \frac{\sum z_i}{4}\right)$.$G = \left(\frac{3+1+3+5}{4}, \frac{2+1+3+1}{4}, \frac{4+0+0+0}{4}\right) = \left(\frac{12}{4}, \frac{7}{4}, \frac{4}{4}\right) = \left(3, \frac{7}{4}, 1\right)$.

Consider the tetrahedron with the vertices $A(3,2,4)$,$B(x_1, y_1, 0)$,$C(x_2, y_2, 0)$,and $D(x_3, y_3, 0)$. If the triangle $BCD$ is formed by the lines $y=x$,$x+y=6$,and $y=1$,then the centroid of the tetrahedron is

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