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

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(C) Given points are $A(2, 3, 5)$,$B(1, 2, 3)$,$C(-5, 4, -2)$,and $D(1, 10, 10)$.First,we find the vectors $\vec{AB}$ and $\vec{CD}$:$\vec{AB} = (1-2,

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$\vec{AB} \parallel \vec{CD}$

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(C) Given points are $A(2, 3, 5)$,$B(1, 2, 3)$,$C(-5, 4, -2)$,and $D(1, 10, 10)$.First,we find the vectors $\vec{AB}$ and $\vec{CD}$:$\vec{AB} = (1-2, 2-3, 3-5) = (-1, -1, -2)$$\vec{CD} = (1-(-5), 10-4, 10-(-2)) = (6, 6, 12)$Now,observe the relationship between $\vec{AB}$ and $\vec{CD}$:$\vec{CD} = (6, 6, 12) = -6(-1, -1, -2) = -6 \vec{AB}$Since $\vec{CD} = k \vec{AB}$ where $k = -6$,the vectors are parallel.Therefore,$\vec{AB} \parallel \vec{CD}$.

If $A(2, 3, 5)$,$B(1, 2, 3)$,$C(-5, 4, -2)$,and $D(1, 10, 10)$,then ...

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