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(D) Let the points be $A, B, C$ with position vectors $a, b, c$ respectively. Since the points are collinear,the vectors $\vec{AB}$ and $\vec{AC}$ are

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$xa + yb + zc = 0, x + y + z = 0$

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(D) Let the points be $A, B, C$ with position vectors $a, b, c$ respectively. Since the points are collinear,the vectors $\vec{AB}$ and $\vec{AC}$ are parallel. Thus,$\vec{AB} = k \vec{AC}$ for some scalar $k$. This implies $(b - a) = k(c - a)$. Rearranging the terms,we get $b - a = kc - ka$,which simplifies to $a(k - 1) + b - kc = 0$. Let $x = k - 1$,$y = 1$,and $z = -k$. Then $xa + yb + zc = 0$. Calculating the sum of the scalars: $x + y + z = (k - 1) + 1 + (-k) = 0$. Therefore,for three collinear points,there exist scalars $x, y, z$ (not all zero) such that $xa + yb + zc = 0$ and $x + y + z = 0$.

If $a, b, c$ are the position vectors of three collinear points,then the existence of scalars $x, y, z$ (not all zero) is such that:

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