MHTCET 18 Apr 2026 Shift 2 Paper

Concept:A plane through three non-collinear points has a normal vector equal to the cross product of two direction vectors lying in the plane.Explanation:Let the three given points be $A(\bar{a}+\bar{b})$, $B(\bar{b}+\bar{c})$ and $C(\bar{a}+\bar{c})$.Using $A$ as the reference point, two direction vectors in the plane are:$B-A = \bar{c}-\bar{a}$$C-A = \bar{c}-\bar{b}$So the normal vector is $(\bar{c}-\bar{a}) \times (\bar{c}-\bar{b})$.Expanding, $(\bar{c}-\bar{a}) \times (\bar{c}-\bar{b}) = \bar{c}\times\bar{c} - \bar{c}\times\bar{b} - \bar{a}\times\bar{c} + \bar{a}\times\bar{b}$.Since $\bar{c}\times\bar{c}=0$ and $\bar{x}\times\bar{y} = -\bar{y}\times\bar{x}$, the normal vector becomes $\bar{a}\times\bar{b} + \bar{b}\times\bar{c} + \bar{c}\times\bar{a}$.Thus the plane equation is $\bar{r}\cdot(\bar{a}\times\bar{b} + \bar{b}\times\bar{c} + \bar{c}\times\bar{a}) = d$.Put $\bar{r} = \bar{a}+\bar{b}$ to find $d$:$d = (\bar{a}+\bar{b})\cdot(\bar{a}\times\bar{b} + \bar{b}\times\bar{c} + \bar{c}\times\bar{a})$.Only two terms contribute: $\bar{a}\cdot(\bar{b}\times\bar{c}) = [\bar{a}\ \bar{b}\ \bar{c}]$ and $\bar{b}\cdot(\bar{c}\times\bar{a}) = [\bar{a}\ \bar{b}\ \bar{c}]$.All other scalar triple products vanish due to repeated vectors.Therefore, $d = 2[\bar{a}\ \bar{b}\ \bar{c}]$.Answer:$\bar{r}\cdot(\bar{a}\times\bar{b} + \bar{b}\times\bar{c} + \bar{c}\times\bar{a}) = 2[\bar{a}\ \bar{b}\ \bar{c}]$, so the correct option is B.