MHTCET 15 May 2024 Shift 2 Paper

Concept:For two lines to intersect, the shortest distance between them must be zero, which means the vector joining a point on each line and the two direction vectors must be coplanar.Explanation:Line 1 passes through $P_1(1, -2, 1)$ with direction vector $\vec{d_1} = (2, 3, 4)$.Line 2 passes through $P_2(3, k, 0)$ with direction vector $\vec{d_2} = (1, 2, 1)$.The vector joining the two points is $P_2 - P_1 = (2, k+2, -1)$.For intersection, the determinant of these three vectors must be $0$.$$\begin{vmatrix} 2 & k+2 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0$$Expanding the determinant:$2(3-8) - (k+2)(2-4) + (-1)(4-3) = 0$$\Rightarrow -10 + 2(k+2) - 1 = 0$$\Rightarrow -10 + 2k + 4 - 1 = 0$$\Rightarrow 2k - 7 = 0$$\Rightarrow k = \frac{7}{2}$Answer:$k = \frac{7}{2}$, so the correct option is A. $\frac{7}{2}$.