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(D) The points are collinear if the area of the triangle formed by them is zero,which implies the determinant is zero: $\begin{vmatrix} 2a & 3a & 1 \

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If $a, 2c/5, b$ are in $H.P.$

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(D) The points are collinear if the area of the triangle formed by them is zero,which implies the determinant is zero: $\begin{vmatrix} 2a & 3a & 1 \ 3b & 2b & 1 \ c & c & 1 \end{vmatrix} = 0$ Expanding the determinant: $2a(2b - c) - 3a(3b - c) + 1(3bc - 2bc) = 0$ $4ab - 2ac - 9ab + 3ac + bc = 0$ $-5ab + ac + bc = 0$ $ac + bc = 5ab$ Dividing by $abc$: $\frac{1}{b} + \frac{1}{a} = \frac{5}{c}$ $\frac{a+b}{ab} = \frac{5}{c} \implies \frac{2ab}{a+b} = \frac{2c}{5}$ This indicates that $\frac{2c}{5}$ is the harmonic mean of $a$ and $b$. Thus,$a, \frac{2c}{5}, b$ are in $H.P.$

When are the points with coordinates $(2a, 3a)$,$(3b, 2b)$,and $(c, c)$ collinear?

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