Which of the following sets of points are non | vedclass

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(C) Points $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ are collinear if the direction ratios of $AB$ are proportional to the directio

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$(-2, 4, -3), (4, -3, -2), (-3, -2, 4)$

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(C) Points $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ are collinear if the direction ratios of $AB$ are proportional to the direction ratios of $BC$.For option $(A)$: Points $(1, -1, 1), (-1, 1, 1), (0, 0, 1)$. Direction ratios of $AB$ are $(-2, 2, 0)$ and $BC$ are $(1, -1, 0)$. Since $(-2)/1 = 2/(-1) = 0/0$ (undefined),these are collinear.For option $(B)$: Points $(1, 2, 3), (3, 2, 1), (2, 2, 2)$. Direction ratios of $AB$ are $(2, 0, -2)$ and $BC$ are $(-1, 0, 1)$. Since $2/(-1) = 0/0 = -2/1$,these are collinear.For option $(C)$: Points $A(-2, 4, -3), B(4, -3, -2), C(-3, -2, 4)$. Direction ratios of $AB$ are $(6, -7, 1)$ and $BC$ are $(-7, 1, 6)$. Since $6/(-7) 
eq -7/1 
eq 1/6$,these points are non-collinear.For option $(D)$: Points $(2, 0, -1), (3, 2, -2), (5, 6, -4)$. Direction ratios of $AB$ are $(1, 2, -1)$ and $BC$ are $(2, 4, -2)$. Since $1/2 = 2/4 = -1/(-2)$,these are collinear.

Which of the following sets of points are non-collinear?

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