MHTCET 11 Apr 2026 Shift 1 Paper

Concept:For perpendicular lines, the dot product of their direction vectors is zero.The plane containing $L_1$ and parallel to $L_2$ has its normal as the cross product of the direction vectors of $L_1$ and $L_2$.Explanation:Direction vector of $L_1$ is $(-3, 2k, 2)$ and of $L_2$ is $(3k, 1, -5)$.Since $L_1 \perp L_2$, their dot product is zero:$(-3)(3k) + (2k)(1) + (2)(-5) = 0$$-9k + 2k - 10 = 0 \Rightarrow -7k - 10 = 0$Hence, $k = -\frac{10}{7}$.Substituting $k$ gives direction vectors:$L_1: \left(-3, -\frac{20}{7}, 2\right)$, which is proportional to $\vec{a} = (-21, -20, 14)$.$L_2: \left(-\frac{30}{7}, 1, -5\right)$, which is proportional to $\vec{b} = (-30, 7, -35)$.The required plane is parallel to both $\vec{a}$ and $\vec{b}$, so its normal is:$\vec{n} = \vec{a} \times \vec{b} = (602, -1155, -747)$.The plane passes through $(1, 2, 3)$ on $L_1$.Thus, the plane equation is:$602(x-1) - 1155(y-2) - 747(z-3) = 0$Expanding and simplifying:$602x - 602 - 1155y + 2310 - 747z + 2241 = 0$$602x - 1155y - 747z + 3949 = 0$.Answer:The correct option is B: $602x - 1155y - 747z + 3949 = 0$.