MHTCET 26 Apr 2025 Shift 2 Paper

Concept:Two lines intersect only if they are coplanar. If the scalar triple product of their direction vectors and the vector joining their known points is non-zero, the lines do not intersect.Explanation:Rewrite each line in standard form.Line 1: $\frac{6x-6}{18}=\frac{y+1}{3}=\frac{z-1}{5}$ simplifies to:$\frac{x-1}{3}=\frac{y+1}{3}=\frac{z-1}{5}$So its direction vector is $\vec{d_1}=(3,3,5)$ and a point on it is $P_1=(1,-1,1)$.Line 2: $\frac{3x+6}{12}=\frac{y-1}{3}=\frac{z+1}{2}$ simplifies to:$\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$So its direction vector is $\vec{d_2}=(4,3,2)$ and a point on it is $P_2=(-2,1,-1)$.Vector joining the two points:$\overrightarrow{P_1P_2}=(-3,2,-2)$Now compute the determinant:$\begin{vmatrix} -3 & 2 & -2 \\ 3 & 3 & 5 \\ 4 & 3 & 2 \end{vmatrix}=61$Since $61 \neq 0$, the lines are not coplanar.Therefore, they do not intersect.Answer:The correct option is C. do not intersect.