MHTCET 25 Apr 2025 Shift 1 Paper

Concept:For two lines to be coplanar, the scalar triple product of the vector joining their fixed points and their two direction vectors must be zero.Explanation:First line has point $P_1(0,2,-3)$ and direction $\vec a_1=(1,2,\lambda)$.Second line has point $P_2(2,6,3)$ and direction $\vec a_2=(2,3,\lambda)$.Coplanarity condition gives:$$\begin{vmatrix} 2 & 4 & 6\\ 1 & 2 & \lambda\\ 2 & 3 & \lambda \end{vmatrix}=0$$Expanding the determinant:$$2(-\lambda)-4(-\lambda)+6(-1)=0$$$$\Rightarrow 2\lambda-6=0$$$$\Rightarrow \lambda=3$$Now, the equation of the plane containing both lines is:$$\begin{vmatrix} x-0 & y-2 & z+3\\ 1 & 2 & 3\\ 2 & 3 & 3 \end{vmatrix}=0$$$$\Rightarrow -3x+3(y-2)-1(z+3)=0$$$$\Rightarrow 3x-3y+z+9=0$$Check the given points in $3x-3y+z+9=0$:A $(1,6,4)$: $3-18+4+9=-2\neq 0$B $(2,8,7)$: $6-24+7+9=-2\neq 0$C $(1,2,3)$: $3-6+3+9=9\neq 0$D $(4,10,9)$: $12-30+9+9=0$So D lies on the plane, while A, B and C do not lie on the plane.Answer:A, B and C do not lie on the plane; D lies on the plane. Hence, as the question asks for the point that does not lie, A, B and C are the non-lying points.