MHTCET 15 Apr 2026 Shift 2 Paper

Concept:Two lines are coplanar when the vector joining one point on each line and the two direction vectors form a scalar triple product equal to zero.Explanation:For the first line, put $s = 0$ to get a point on it.$A(-1, 3, 1)$ lies on the line, and its direction ratios are $(1, -\lambda, \lambda)$.For the second line, put $t = 0$ to get a point on it.$B(0, 1, 2)$ lies on the line, and its direction ratios are $\left(\frac{1}{2}, 1, -1\right)$.The vector joining A and B is:$$\overrightarrow{AB} = B - A = (0+1, 1-3, 2-1) = (1, -2, 1)$$For coplanarity, the scalar triple product of $\overrightarrow{AB}$ and the two direction vectors must be zero:$$\begin{vmatrix} 1 & -2 & 1 \\ 1 & -\lambda & \lambda \\ \frac{1}{2} & 1 & -1 \end{vmatrix} = 0$$Expanding along the first row:$$1(\lambda - \lambda) + 2\left(-1 - \frac{\lambda}{2}\right) + \left(1 + \frac{\lambda}{2}\right) = 0$$$$0 - 2 - \lambda + 1 + \frac{\lambda}{2} = 0$$$$-1 - \frac{\lambda}{2} = 0$$$$\lambda = -2$$Answer:$\lambda = -2$, which matches Option C.