MHTCET 15 Apr 2026 Shift 1 Paper

Concept:Two lines are coplanar only if the vector joining their fixed points and their two direction vectors are coplanar, so their scalar triple product must be zero.Explanation:The first line passes through $A(1,-1,0)$ and has direction vector $\vec{d_1}=(2,\lambda,2)$.The second line passes through $B(-1,-1,0)$ and has direction vector $\vec{d_2}=(5,2,\lambda)$.The vector joining the fixed points is:$$\overrightarrow{AB}=B-A=(-2,0,0)$$For coplanarity, the scalar triple product is zero:$$\begin{vmatrix} -2 & 0 & 0\\ 2 & \lambda & 2\\ 5 & 2 & \lambda \end{vmatrix}=0$$Expanding along the first row:$$-2(\lambda^2-4)=0$$$$\lambda^2=4$$$$\lambda=\pm 2$$For $\lambda=2$, the direction vector is $(2,2,2)$.A normal to the plane is $\vec{d_1}\times\overrightarrow{AB}=(0,-4,4)$, which gives the plane:$$y-z+1=0$$For $\lambda=-2$, the direction vector is $(2,-2,2)$.A normal to the plane is $(0,-4,-4)$, which gives the plane:$$y+z+1=0$$Combining both cases, the required plane is:$$y\pm z+1=0$$Answer:$y\pm z+1=0$, i.e. Option B.