p points are chosen on each of the three copl | vedclass

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(D) Total number of points is $3p$. To form a triangle,we need to select $3$ points that are not collinear. The total number of ways to select $3$ poi

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$p^2(4 p-3)$

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(D) Total number of points is $3p$. To form a triangle,we need to select $3$ points that are not collinear. The total number of ways to select $3$ points from $3p$ points is $^{3p}C_3$. Since $p$ points are collinear on each of the $3$ lines,we must subtract the sets of $3$ points that lie on the same line. Number of triangles $= ^{3p}C_3 - 3 	imes (^pC_3)$. $= \frac{3p(3p-1)(3p-2)}{6} - 3 	imes \frac{p(p-1)(p-2)}{6}$. $= \frac{p(3p-1)(3p-2) - p(p-1)(p-2)}{2}$. $= \frac{p}{2} [ (9p^2 - 9p + 2) - (p^2 - 3p + 2) ]$. $= \frac{p}{2} [ 8p^2 - 6p ] = p(4p^2 - 3p) = p^2(4p-3)$.

$p$ points are chosen on each of the three coplanar lines. The maximum number of triangles formed with vertices at these points is

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