P_1 and P_2 are two distinct and intersecting | vedclass

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(B) tetrahedron is formed by $4$ non-coplanar points.Let the set of $3$ points on $P_1$ be $S_1 = \{A, B, C\}$ and the set of $3$ points on $P_2$ be $

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$15$

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(B) tetrahedron is formed by $4$ non-coplanar points.Let the set of $3$ points on $P_1$ be $S_1 = \{A, B, C\}$ and the set of $3$ points on $P_2$ be $S_2 = \{D, E, F\}$.Since $S_1$ are coplanar (on $P_1$) and $S_2$ are coplanar (on $P_2$),we cannot choose $4$ points from $S_1$ or $4$ points from $S_2$.To form a tetrahedron,we must select points such that they are not all on the same plane.Case $1$: Select $3$ points from $S_1$ and $1$ point from $S_2$.Number of ways $= {}^3C_3 	imes {}^3C_1 = 1 	imes 3 = 3$.Case $2$: Select $2$ points from $S_1$ and $2$ points from $S_2$.Number of ways $= {}^3C_2 	imes {}^3C_2 = 3 	imes 3 = 9$.Case $3$: Select $1$ point from $S_1$ and $3$ points from $S_2$.Number of ways $= {}^3C_1 	imes {}^3C_3 = 3 	imes 1 = 3$.Total number of tetrahedrons $= 3 + 9 + 3 = 15$.

$P_1$ and $P_2$ are two distinct and intersecting planes. Three non-collinear points lie on $P_1$ and another three non-collinear points lie on $P_2$ (none being on the line of intersection of the planes). Then,the maximum number of tetrahedrons formed using these six points is:

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