If A(2, 4, -1),B(3, 6, -1),and C(4, 5, 1) are | vedclass

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(D) Let the vertices of the parallelogram be $A(2, 4, -1)$,$B(3, 6, -1)$,$C(4, 5, 1)$,and $D(x, y, z)$.Since the diagonals of a parallelogram bisect e

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$(3, 3, 1)$

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(D) Let the vertices of the parallelogram be $A(2, 4, -1)$,$B(3, 6, -1)$,$C(4, 5, 1)$,and $D(x, y, z)$.Since the diagonals of a parallelogram bisect each other,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.Midpoint of $AC = \left( \frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2} \right) = \left( 3, \frac{9}{2}, 0 \right)$.Midpoint of $BD = \left( \frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2} \right)$.Equating the midpoints:$\frac{3+x}{2} = 3 \implies 3+x = 6 \implies x = 3$.$\frac{6+y}{2} = \frac{9}{2} \implies 6+y = 9 \implies y = 3$.$\frac{-1+z}{2} = 0 \implies -1+z = 0 \implies z = 1$.Thus,the fourth vertex $D$ is $(3, 3, 1)$.Therefore,option $(D)$ is correct.

If $A(2, 4, -1)$,$B(3, 6, -1)$,and $C(4, 5, 1)$ are three consecutive vertices of a parallelogram,then its fourth vertex is

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